Integration Bee Topics Guide for
Problem Writers
Yuepeng Alex Yang
2nd Edition
I am very thankful to a lot of people. To my family for letting me pursue my dreams,
my best friend for getting me into speed integration, and to the organizers of MUSA of UC
Berkeley, Berkeley Math Tournament, and Stanford Math Tournament for accepting me as a
problem writer for Integration Bees. This book would never exist without them.
1
Contents
1 Qualifying Round Topics
1.1 Basic Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Basic U-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Basic Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Basic Trig Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Basic Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Basic Trig-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 Basic Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9 Area of a Partial Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.10 Algebra Manipulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.11 Intermediate U-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.12 Absolute Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.13 Reverse Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.14 Basic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.15 Intermediate Trig Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.16 Intermediate Trig-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.17 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.18 King’s/Queen’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.19 Gaussian Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.20 Special Wildcard . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
7
8
9
11
12
13
14
15
16
18
20
21
22
24
26
27
29
31
35
37
2 Regular Round Topics
38
2.1 Modified Textbook Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.2 Sneaky Algebra Manipulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.3 Basic Forced U-Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.4 Gamma Function Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.5 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.6 Trig-Identities Manipulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
2.7 Intermediate Trig-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2.8 Infinite Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
2.9 Sneaky Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2.10 Sneaky Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2
3 Intermediate Topics
52
3.1 Power Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.2 Sneaky U-Sub Tricks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.3 Wallis’ Trick . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.4 Advanced Trig-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.5 Awkward & Advanced Integration by Parts . . . . . . . . . . . . . . . . . . . 62
3.6 Intermediate Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.7 Basic Periodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.8 Speed Bashing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4 A Little Advanced Topics
69
4.1 Undertow Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.2 Square-Square Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3 1 + sin(2x) Trick . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.4 Advanced King’s/Queen’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 75
4.5 Weierstrass Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
4.6 Reverse Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
5 Advanced Topics
83
5.1 Semi-Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
5.2 Advanced Trig Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
5.3 Advanced Gaussian Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
5.4 Dirichlet Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
6 Very Advanced Topics
92
6.1 Inverse Jailbreaking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
6.2 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
6.3 Complexifying the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
6.4 Ninja Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
6.5 Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
6.6 Frullani Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
6.7 Feynman Technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
6.8 Modified Famous Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
6.9 Unique Tricks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
3
Quality Control
Before you get crazy writing integrals for the integration bee, let’s first get some
things straight before you end up becoming high-school me. Every integral problem must be
clean. No messy answers and no tedious process. Let me give you some examples.
A Very Long Answer
Z
tan−1 (x2 )dx
[UCLA Homework]
Yes, I get it, this is a cool looking integral. But the issue is, the answer is not clean
at all. I mean...look at this:
Z
Z
2x2
−1 2
−1 2
tan (x )dx = x tan (x ) −
dx
x4 + 1
I’m not even gonna evaluate it, I know it’s a long answer, and you don’t want competitors
writing long answers. Imagine having to check if they’re actually correct; looking at every
single detail:
!
√
2
√
√
1
x
+
1
1
2x
+
1
√
x tan−1 (x2 ) + √ ln
+ √ tan−1 (1 − 2x) − √ tan−1 (1 + 2x) + C
2 2
x2 − 2x + 1
2
2
Remember that not everyone has good handwriting. Yea no, we’re not having this.
A Very Scary Long Process
Z 1
sin−1 (x) ln(x)dx = 2 −
0
π
− ln(2)
2
Let’s say you used WolframAlpha and you noticed that the indefinite integral was
long. So you added bounds to make the answer shorter. You used WolframAlpha to test
which bounds make the cleanest answer. You ended up getting this. Okay, yes, it’s clean, but
is the process clean??? Please try it out yourself before inconsiderately putting this on the
“earlier rounds”.
√
Z 1
h
i1 Z 1
√
1 − x2
sin−1 (x) ln(x)dx = ln(x)(x sin−1 (x) + 1 − x2 ) −
sin−1 (x) +
dx
x
0
0
0
π
= 1 − lim x ln(x) sin (x) + 1 − −
x→0
2
−1
Z π
2
0
cos2 θ
dθ
sin θ
As you can see, there are some tedious issues with this problem being in the earlier rounds.
Realistically, competitors will hesitate and get disgusted in the middle of the integration by
parts process, there’s a tedious trig-sub they need to evaluate, and an annoying limit.
4
A “Made in China” Integral
Z 1
(1 + x + 1 + 11 + x + 11x − x − 111x − 1010x)dx
0
[MIT Integration Bee 2015 - Round 1 Problem 6]
Can we not??? This is just straight up rude. I’m here to integrate, not do arithmetic.
This is a poor quality integral. Fun to whoever wrote this, but not fun who’s actually
competing.
Annoying Constants
Z
sec2 (x)
dx
sec2 (x − 1)
[My High School Integration Bee Journal]
Honestly, I don’t even know how I came up with this. All I know is that there were
so many annoying constants.
Z
Z
sec2 (x)
2
dx = cos (x − 1) tan(x) − sin(2x − 2) tan(x)dx
sec2 (x − 1)
Z
⇒ 2 sin2 (x) cos(2) − sin(2) sin(x)(2 cos(x) − sec(x))dx
Because of those sin(2) and cos(2) constants, I completely lost motivation to continue solving
further. This is also a long answer and even with bounds, WolframAlpha would give me this:
Z 1
sec2 (x)
dx = cos(2) + sin2 (1) tan(1) − sin(2) ln(cos(1))
2
0 sec (x − 1)
Notice that I could’ve gotten a different correct answer. But how would you know that???
Also, this integral is cruel:
Z 1
sec2 (x)
sec2 (x − 1)
−
dx = 0
2
sec2 (x)
0 sec (x − 1)
I would avoid integrals that do equal to zero. Because some competitors can mentally graph
weird functions and guess zero; it makes this integral pointless and unfair for the other
opponent who is actually integrating.
5
Caterpillar U-Substitutions
3
3
x2 ex cos(ln(ex + 1))
Z
(ex3 + 1)
p
dx
sin(ln(ex3 + 1))(sin(ln(ex3 + 1)) + 1)
[My High School Integration Bee Journal]
If I was a competitor and I saw this, my eyes will be blinded from all the messiness
of symbols and functions.√ Let u = x3 . Then let w = eu + 1. Then let y = ln(w). Then let
t = sin(y). Then let r = t.
Z
2
dr
2
3
r +1
Seriously? It took me 5 u-substitutions to get here. Let’s not overdo this please...
A Textbook Problem
Z
x−1
√
dx
2x2 − 3
[Caltech-Harvey Mudd Math Competition - Finals Round Problem 40]
Let’s also not force competitors to bash. If you’re gonna bash the competitors, you
better do it with style. This integral is not one of them. Also, throwing a textbok problem
for finals round is pure laziness. Let’s not do this.
√
√
√
Z
2x2 − 3 ln | 2x2 − 3 + 2x|
x−1
√
√
dx =
−
+C
2
2x2 − 3
2
This literally looks like an integral table information. Terrible.
Force Graphing
Z 10
|4 − |3 − |2 − |1 − |x|||||dx
−10
[Carnegie Mellon Integration Bee 2023 - Qualifying Exam]
This is very rude. I’m here to integrate, not graph and do geometry!! I get it, you
want to be conceptual with integrals, but this isn’t MIT. The amount of trapezoidal areas I
have to add up. (Vomit emoji here.) I highly doubt anyone would mentally graph this. If so,
I’m very concerned.
Z 10
Z 9
2
|4 − |3 − |2 − |1 − x||||dx = 2
|4 − |3 − |2 − |u||||du
−1
0
Z 1
Z 9
(3 − u)du + 2
=4
0
|4 − |3 − |2 − u|||du
1
I think I’ll pass on this...
6
Chapter 1
Qualifying Round Topics
In this round, competitors are taking an exam trying to solve 20 integrals in one
hour (usually). At least the first 5-6 integrals should be easy for competitors to integrate,
then increase difficulty drastically so that we can easily choose our top 16 competitors.
1.1
Basic Integration
I think textbooks have good basic integration problems for this, but we can definitely
make it more stylish or satisfying. Here are some good examples.
Textbook Problems:
1.
4.
Z
Z
2
(5x − 8x + 5)dx
2.
Z
x
4 sin
5.
Z
6.
Z
−x
(e + x + e )dx
3.
Z
√
1+ x
dx
x
x
3
dx
4
dx
1 + x2
√
7
dx
1 − x2
Modified Problems:
1.
4.
Z
Z
2
(x − 1)(x + 1)(x + 1)dx
5.
2.
Z
sec(x) + tan(x)
dx
cos(x)
Z 2
x +2
dx
x2 + 1
(ex + 1)3 dx
6.
3.
Z
1
√
x
√
1
√ + 1 + x dx
x
Z
7
1
√
x+1
√
1
x+1+ √
1−x
dx
1.2
Basic U-Substitution
Here are some good examples of basic u-substitution integrals for qualifying exam:
1.
Z
2.
11.
x
dx
x2 + 1
Z
1
√
dx
3
x − 2024
12.
Z
ex
dx
e2x + 1
Z
2
tan(x) sec (x)dx
3.
13.
Z
x
x
Z
cos(x) + 1 + ex
dx
sin(x) + x + ex
Z
sin(x) cos(x)
dx
sin2 (x) + 1
(e + 1)(e + x)dx
4.
14.
Z
sin(sin(x) + 1) cos(x)dx
5.
6.
Z
16.
Z
x+ex
e
7.
15.
√
x x − 1 x + 1dx
√
Z
Z
dx
17.
Z
tan(x)dx
9.
18.
Z
Z
1
x
1
ln(x) +
ln(x)
Z √ x √ −x
e − e
√
dx
√
ex + e−x
dx
19.
Z 2
(x + x + 1)(2x + 1)dx
10.
20.
Z
(x + 1)
2024
sin(ln(x))
dx
x
Z
2
sec (5x − 4)dx
8.
2
(x + 1)ex +2x dx
1
1+
x
ln(xex )dx
Z
csc(1 − x) cot(1 − x)dx
dx
8
1.3
Symmetry
Symmetry Concept
For any real a, if f (x) is an odd function, then
Z a
f (x)dx = 0.
−a
If f (x) is an even function, then
Z a
Z a
f (x)dx = 2
−a
f (x)dx.
0
Example 1
Z 1
(x + 1)(x2 + 1)dx
−1
Z 1
3
Z 1
2
(x + x + x + 1)dx = 2
−1
(x2 + 1)dx =
0
8
3
Example 2
Z 1
x
−1
sin−1 (x)
1
+
−1
tan (x) x
dx
The left term is an odd function, so the answer is just 2.
Example 3
Z 6
0
Let u = x − 3.
Z 3
(x − 2)(x − 3)(x − 4) + 1
dx
x2 − 6x + 10
u(u2 − 1) + 1
du =
u2 + 1
−3
Z 3
du
−3
u2 + 1
9
= 2 tan−1 (3)
Here are some odd functions to give you some inspirations. All of these integrals
should equal to zero.
1.
Z 1
11.
√
√
√
( 7 x + 5 x + 3 x)dx
Z 1
−1
12.
2.
Z 1
sin−1 (tan−1 (x))dx
−1
Z 1
x + sin(x)
dx
−1 1 + cos(x)
sin−1 (x) + tan−1 (x)dx
−1
13.
3.
Z 1
1 − cos(x)
dx
x
−1
Z 1
1
1
−1
+ tan
x−
dx
x+
x
x
−1
tan
−1
4.
Z 1
ln(x +
√
x2 + 1)dx
14.
Z 1
ln
−1
1 + sin(x)
1 − sin(x)
Z 1
−1
5.
Z 1
15.
(ex − e−x )dx
ln
−1
−1
6.
Z 1
16.
sec(x) tan(x)dx
Z 1
ln
−1
ex + 1
e−x + 1
dx
x3 + x
dx
4
2
−1 x + x + 1
17.
dx
Z 1
−1
7.
1+x
1−x
Z 1
dx
2
sin(x)ex dx
−1
8.
18.
Z 1
Z 1
sin(sin(x))dx
−1
−1
9.
19.
Z 1
Z 1
sin(x3 + x) + sin−1 (x3 + x)dx
Z x
f (x)dx where f (x) =
−1
10.
(cos(x))x − (sec(x))x dx
−1
20.
Z 1r
q
√
3
3
x + x + 3 x dx
−1
0
Z 1
−1
10
√
1−x−
√
2
e−t dt
1 + x dx
1.4
Basic Exponential Functions
Here are some good basic examples of exponential integrals for the qualifying exam:
1.
2.
3.
Z
ex
(ex + 1)2
Z ∞
2x + 3x
dx
5x
0
6.
7.
Z
11.
12.
Z
ex + e2x
dx
1 + e2x
(ex + 1)(e−x + 1)dx
Z
14.
e +1
dx
x + ex + 1
Z
dx
ex
dx
(ex − 1) ln(ex − 1)
15.
Z
(ex + e−x )4 dx
Z
Z
Z 16.
x−1 x+1
e
Z
17.
x
Z
18.
dx
x
Z ∞
0
11
2
Z q
√
ex ex dx
x
e7x + e4x
dx
e5x + e2x
ex
cos(e2x )
ee (e2x + ex )dx
dx
e sin(e )dx
9.
ex
dx
1 − e2x
ex + e−x
x
Z
√
Z
dx
ex + 1
2
8.
Z
13.
4.
5.
10.
ex
dx
ex + 2024
Z
1+
√
1 − e−x
dx
ex
1.5
Basic Trig Integrals
Here are some good basic examples of trigonometric integrals for the qualifying exam:
1.
2.
Z
11.
sin(x) + cos(x)
dx
cos(x)
Z
12.
Z
tan(x) + cot(x)
dx
sin(x) cos(x)
Z
(sin(x) + cos(x))2 dx
2
sin (x) cos(x)dx
3.
4.
5.
13.
Z
Z
sin2 (x) cos3 (x)dx
14.
Z
Z
(sin(x) + 1)(cos(x) + 1)dx
(tan(x) sec(x))2 dx
15.
Z
Z
tan(x) sec3 (x)dx
16.
6.
Z
(tan(x) + sec(x))2 dx
(sec(tan(x)) sec(x))2 dx
Z
tan2 (x)dx
17.
7.
Z
Z
cos(x)
dx
sin2 (x) + 1
18.
8.
9.
10.
Z
Z
csc(x) cot(x)
dx
1 + csc(x)
Z
tan(x) + 1
dx
sec(x)
(1 + sin(x))(1 + sec2 (x))dx
19.
Z
sin(x) − cos(x)
dx
sin(x) + cos(x)
Z
1
dx
sec(x) + tan(x)
20.
3
sin (x)dx
12
(sec(x) − csc(x))(sec(x) + csc(x))
dx
tan(x) + cot(x)
Z
sin(x) + cos(x)
dx
sec(x) + csc(x)
1.6
Basic Integration by Parts
Here are some good basic examples of integration by parts integrals for the qualifying
exam. Notice that there are no recursive integration by parts problems.
1.
11.
Z
Z
2 x
x e dx
2.
12.
Z
(x + 1)2 cos(x)dx
Z
tan−1 (x)dx
Z
√
ln(x)dx
3.
13.
Z
x sin(x)dx
4.
Z
5.
14.
x
dx
ex
15.
Z
2
Z
ln(x2 + 1)
dx
x2
Z
sin−1 (x)dx
x sec (x)dx
6.
16.
Z
Z
x ln(x)dx
7.
Z
17.
x sec2 (x) tan(x)dx
Z
x 2
(xe ) dx
8.
18.
Z
x 2
10.
Z
x2x dx
Z
x(cos(x) + sin(x))dx
(x + e ) dx
9.
x ln(x)dx
19.
ln(x)
dx
x2
20.
Z
sin(x) ln(sec(x))dx
13
Z p
ex (x2 + 6x + 9)dx
Z
ln(x)
√ dx
x
1.7
Basic Trig-Substitution
Here are some good basic examples of trig-sub integrals for the qualifying exam:
1.
Z
2.
Z
dx
√
x2 x2 − 1
3.
Z
dx
(1 − x2 )
4.
Z
6.
dx
√
x x2 − 1
7.
8.
Z √
1 − x2
dx
x2
x2
Z
3
2
3
(1 − x2 ) 2
9.
dx
3
Z
dx
x2 + 4
(1 + x2 ) 2
5.
Z
10.
dx
3
(x2 − 1) 2
Z
√
dx
1
dx
9 − x2
Z √ 2
x −1
dx
x
To make an intentional trig-sub integral, start with a skeleton. For example:
Z
dθ
sec θ + tan θ
If I let x = sin θ, I need dx = cos θdθ:
Z
Z
Z
cos θ
dx
dx
dθ =
=
(sec θ + tan θ) cos θ
1 + sin θ
1+x
Too easy, that’s just basic substitution.
If I let x = sec θ, I need dx = sec θ tan θdθ:
Z
Z
sec θ tan θdθ
dx
√
√
=
2
(sec θ + tan θ) sec θ tan θ
(x + x − 1)x x2 − 1
This looks good, let’s save this!!
√
If I let x = tan θ, I need dx = sec2 θdθ: note that sec(tan−1 (x)) = 1 + x2 .
Z
Z
sec2 θdθ
dx
√
=
2
(sec θ + tan θ) sec θ
( 1 + x2 + x)(1 + x2 )
This also looks good as well!! Although the answer may be long, we can just add bounds
to make answer cleaner. In this case, I would go from 0 to ∞ so that my skeleton goes
from tan−1 (0) = 0 to tan−1 (∞) = π2 (since x = tan θ).
14
1.8
Basic Partial Fractions
Here are some good basic examples of partial fraction integrals for the qualifying
exam:
1.
2.
3.
Z
4.
10.
Z
dx
x(x + 1)
Z
dx
2
x (x + 1)
11.
dx
(x + 1)(x + 2)
12.
Z
5.
dx
x(2x − 1)
Z
dx
Z
x+2
x2 + 4x + 3
7.
8.
9.
Z
14.
15.
Z
e2x
dx
e2x + 3ex + 2
Z
x2 − x
dx
1 + x + x2 + x3
Z
3x
dx
x2 + 2x − 8
Z
dx
3
x −x
x+5
x2 + x − 2
x
dx
(x + 1)(x + 2)(x + 3)
Z
(x + 1)(x + 4)
dx
(x + 2)(x + 3)
dx
16.
Z
dx
x(1 + x)2
Z
13.
x2 − 1
6.
Z
17.
dx
2
x (1 + x2 )
Z 4
x−4
2
x2 − 6x + 5
Z ∞
3
15
dx
dx
dx
(x − 2)(x + 6)
1.9
Area of a Partial Circle
Area of a Partial Circle
√
We are familiar with the graph of a semicircle: y = r2 − x2 where r is the radius of
the semicircle. Then,
Z r√
πr2
r2 − x2 dx =
2
−r
Z r√
Z 0√
πr2
r2 − x2 dx =
r2 − x2 dx =
4
0
−r
Basic Example
Z 1√
1 − x2 dx
0
We see that this is a quarter of a circle of radius 1. So the integral is the area π4 .
Hidden Symmetry
Z 2
x 1 √
3
+
4 − x2 dx
x cos
2
2
−2
[Free Wi-Fi Password]
To most people this looks intimidating, but it’s actually easy if you know the concept
of symmetry and the partial area of a circle. Notice that the left term is an odd function:
Z 2
x √
3
x cos
4 − x2 dx = 0
2
−2
In the right term, we can use symmetry of an even function:
Z
Z 2√
1 2√
2
4 − x dx =
4 − x2 dx = π
2 −2
0
Ellipse Example
Z 2√
12 − 3x2 dx
0
[MIT Integration Bee 2013 - Qualifying Exam]
We need that x2 to not have a coefficient other than 1.
Z 2√ √
√
3 4 − x2 dx = 3π
0
16
Substitution Example
Z 3√
6x − x2 dx
0
Let’s complete the square and perform u-sub.
Z 3p
Z 0√
9π
2
9 − (x − 3) dx =
9 − u2 du =
4
0
−3
Exponential Example
Z ∞√
e−2x − e−4x dx
0
Z ∞
e
−x
√
1 − e−2x dx =
Z 1√
0
1 − u2 du =
0
π
4
Hybrid Example
Z 1
−1
Z 1
−1
2 − x2
√
1 − x2
1 + x3 dx
√
1
2
√
+ 1−x
1 + x3 dx
1 + x dx =
1 − x2
−1
Z 1
√
1
π
3π
2
√
+ 1 − x dx = π + =
=
2
2
2
1−x
−1
2 − x2
√
1 − x2
3
Z 1
Hopefully this was easy to follow along. We separate it in a way where the radicals are easy
to understand. Then we used symmetry to get rid of x3 . Then the rest of the integral is just
inverse sine and the partial area of a circle.
17
1.10
Algebra Manipulation
Zero Substitution
Z
x2
dx =
x2 + 1
Z
x2 + (1 − 1)
dx =
x2 + 1
Z
1−
1
x2 + 1
dx
It may not seem much, but this is actually the #1 most helpful algebra trick for speed
integration!! Here, try this one. It’s not always obvious sometimes.
Z
dx
2
(x + 2) (x2 + 4x + 5)
[JEE Main Training]
Hint: Let 1 = 5 − 4 = (x2 + 4x + 5) − (x2 + 4x + 4).
Geometric Polynomial Division
Z
Z 4
Z
x4
x −1+1
1
dx =
dx = x3 − x2 + x − 1 +
dx
x+1
x+1
x+1
Z
x7
dx =
x2 − 1
Z
x7 − x + x
dx =
x2 − 1
Z
6
x −1
x
x
dx
+ 2
2
x −1
x −1
Forcing an Arctangent
We mainly complete the square to give ourselves an arctan form.
Z
Z
2
dx
dx
2x + 1
−1
√
=
= √ tan
+C
x2 + x + 1
(x + 21 )2 + 34
3
3
du the Numerator
We purposely put down the derivative of the denominator on the numerator so that it’s
an easy u-sub. Then we just deal with the leftovers.
Z
Z
5x2 + 3x − 2
3x2 + 4x 2x2 − x − 2
dx
=
+
dx
x3 + 2x2
x3 + 2x2
x3 + 2x2
Z
Z
3x2 + 4x 2x2 − (x + 2)
3x2 + 4x
2
1
=
+
dx
=
+
− 2 dx
3
2
2
3
2
x + 2x
x (x + 2)
x + 2x
x+2 x
18
Function Conjugating
We multiply top and bottom with functions to get rid of fractions or other types of
simplification.
Z
Z
Z
Z
2ex + 2
ex
2
2(1 + ex )
dx
=
dx
=
+ 1 dx
dx
=
1
(ex + 1) + 1
ex + 2
ex + 2
1 + 1+e
x
Z
dx
=
(4 cos(x) + sin(x))2
Z
sec2 (x)
=
((4 cos(x) + sin(x)) sec(x))2
Z
sec2 (x)
dx
(4 + tan(x))2
Radical Conjugating
Z
x
√
dx =
2
x +1−x
Z
Fake Partial Fractions
Z
Z
√
Z √
x
( x2 + 1 + x)
√
√
dx = (x x2 + 1 + x2 )dx
2
2
( x + 1 − x) ( x + 1 + x)
dx
√
√
=
x+1− x−1
Z
(x + 3)
dx =
(x + 2)2 (x + 4)2
√
√
( x + 1 + x − 1)dx
Z
x+3
(x2 + 6x + 8)2
dx
This is just a basic u-substitution.
Z
(x + 1)2
dx =
(x + 2)(x2 + x + 1)
Z
(x + 1)2
dx =
x3 + 3x2 + 3x + 2
Z
(x + 1)2
dx
(x + 1)3 + 1
This one is a little tricky to see, but you can simply do u = (x + 1)3 + 1.
19
1.11
Intermediate U-Substitution
These integrals are u-substitutions that are a little hard to see, but also not too
difficult. Here are some good examples.
1.
Z
11.
x
dx
1 + x4
12.
2.
Z
Z
dx
√
x(1 + x)
x+5
p
dx
(x + 2)(x + 8)
13.
3.
Z
3
√
x2 x − 1dx
Z
e2x
dx
1 + e4x
Z
x2
x e dx
14.
4.
Z
5.
1
15.
Z
x2
dx
(x − 2)3
p
sin(x)
p
dx
2
cos (x) cos(x)
16.
6.
Z
7.
9.
2
√
cos( x)
√
dx
x
Z
8.
Z 3
x
√
dx
x−1
1 1
e x dx
x2
Z
√
Z
Z 1
Z
sec(x) ln(sec(x) + tan(x))dx
dx
x − x2
17.
√
x sin(x x)dx
18.
√
19.
2
7
x (2x − 1) dx
Z
√
Z
Z
1
x5
r
20.
x4 + 1
dx
x4
20
sec2 (x)
dx
8 + sec2 (x)
Z r
0
10.
dx
√
x(1 + 2 x + x)
Z
x
dx
1 − x3
ln(x) + 1
dx
x ln(x) + 1
1.12
Absolute Values
Here are some good basic examples of absolute value integrals for the qualifying
exam:
1.
6.
Z 2
2
x − 1 dx
Z 4
−2
2.
Z 5
0
3.
0
7.
Z ∞
dx
x + |x − 1|
Z 1
8.
Z e
|ex − 1| dx
1
9.
Z 1
x+1
dx
|x − ln(x)| + ln(x)
Z 16
|x − |x − |x|||dx
√
x − 2 dx
0
−1
5.
e−|x| dx
−∞
−1
4.
|x − 1|
dx
|x − 2| + |x − 3|
10.
Z 2π
Z 2
| sin(x)|dx
−2
−2π
dx
p
|1 − x2 | + 1
WARNING EXAMPLE
Z 2π q
Z 2π
2
| cos(x)|dx = 4
1 − sin (x)dx =
0
0
Not only this tricks competitors, but nearly everyone when it comes to square roots. Just be
very careful with this, but also a nice way to give a hidden absolute value integral.
21
1.13
Reverse Product Rule
Reverse Product Rule Concept
Z
[f ′ (x)g(x) + f (x)g ′ (x)] dx = f (x)g(x) + C
As simple as it looks, it’s not always clear to see in most integration bee problems.
An Obvious Example
Z
sin(x)
+ cos(x) ln(x)dx
x
If I didn’t tell you this concept, you might have struggled solving this integral. But
now that I showed you this trick, you are aware that this is just a reverse product rule. So
the answer is clearly sin(x) ln(x) + C.
Exponential Example
Z
ex (sin(x) + cos(x))dx
To most beginners, they would probably do integration by parts. But if you know
the trick, you could definitely solve this in seconds. In fact, it’s good to keep in mind:
Z
ex (f (x) + f ′ (x))dx = ex f (x) + C
A Trig Example
Z
Z
=
sec(x)(2x + x2 tan(x))dx
[2x] sec(x) + x2 [sec(x) tan(x)]dx = x2 sec(x) + C
22
Logarithm Example
Z
(ln2 (x) + 2 ln(x))dx
Natural logs are very good at hiding xn due to the cancellation from the derivative
of ln(x).
Z
1 · ln2 (x) + x ·
2 ln(x)
dx = x ln2 |x| + C
x
In fact,
Z
[f (ln(x)) + f ′ (ln(x))] dx = xf (ln(x)) + C
Polynomial Example
Z 0
x(x2 + 1)2 (3x + 1)2 + (3x + 1)(x2 + 1)3 dx
− 13
[WolframAlpha Suggestions]
You’re probably guessing that the indefinite answer is (x2 + 1)3 (3x + 1)2 . Well you’re
wrong lol. You must be aware of removed constants; this tricks competitors who just likes
to speed guess. The indefinite answer is 61 (x2 + 1)3 (3x + 1)2 . So the definite integral answer
should just be 61 .
23
1.14
Basic Series
Some Basic Series Formulas
The following properties are true if |x| < 1.
1.
6.
∞
X
xn =
n=0
1
1−x
∞
X
(−1)n
n=0
2.
∞
X
x2n+1
= tan−1 (x)
2n + 1
7.
x
nx =
(1 − x)2
n=1
n
∞
2n+1
X
n x
(−1)
= sin(x)
(2n + 1)!
n=0
3.
∞
X
n
n=1
xn
=
x
(1 − x)2
8.
∞
X
x2n
(−1)n
= cos(x)
(2n)!
n=0
4.
∞
X
xn
n=0
n!
9.
= ex
1
=1
n(n + 1)
n=1
5.
∞
X
xn
n=1
n
∞
X
10.
∞
X
1
= − ln (1 − x)
n=1
n
=
2
π2
6
Geometric Series Example
Z 1√
1 + x2 + x4 + x6 + ... dx
0
v
Z 1r
Z 1u
∞
uX
t
=
x2n dx =
0
0
n=0
1
π
dx =
2
1−x
2
Out of Convergence Example
Z 2024
dx
(1 + x2 )(1 + x4 )(1 + x8 )...
0
Z 1
Z 1
(1 + x)
2
=
dx =
(1 − x2 )dx =
2
3
4
3
0 1 + x + x + x + x + ...
0
Note that the integral with bounds [1, 2024] is out of convergence for the geometric
series. Because of this, the infinite denominator goes off to infinity and makes the
integral equal to zero.
24
Taylor Series Example
s
Z
2x
e
r
ex2
3
e x3
q
√
4
5
ex4 ex5 · · · dx
[Silver]
Z
=
x
e e
xn
n=1 n!
P∞
Z
dx =
x
x
ex ee −1 dx = ee −1 + C
Numerical Series Example
v s
r q
Z u
u
√
t x
x
2 4 8x 16x ...dx
[Silver]
Z
=
2
x
+ 2x
+ 3x
+ 4x
+...
2
4
8
16
Z
P
n
x( ∞
n=1 2n )
dx =
2
Z
dx =
22x dx
Composite Function Series Example
√
√
x 3
5
ee ee3x ee5x
√ 2x √
√
...dx
4
6
ee
ee4x ee6x
Z
Z P
−(−ex )n
∞
x
= e n=1 n dx = eln(1+e ) dx = x + ex + C
Z
Telescoping Series Example
Z q
q
q
√
√
√
4 √
6 √
3
5
3
5
x x·
x x·
x 7 x...dx
Z
=
x
1
1
1
1
+ 2(3)
+ 3(4)
+ 4(5)
+...
2
Z
dx =
x
P∞
1
n=1 n(n+1)
dx =
x2
+C
2
Gaussian Series Example
√
√
√
Z
ln(x) ln( x) ln( 3 x) ln( 4 x)
+
+
+
...dx
x
2x
3x
4x
Z
ln(x) ln(x) ln(x) ln(x)
=
+
+
+
...dx
x
4x
9x
16x
Z
∞
ln(x) X 1
π 2 ln2 (x)
=
dx
=
+C
x n=1 n2
12
25
1.15
Intermediate Trig Integrals
In this section, we start utilizing trig-identities, including some uncommon ones.
Here are some good examples for the qualifying exam:
1.
Z π
2
12.
2
sin (x)dx
Z
sin(x − sin(x)) − sin(x + sin(x))dx
0
2.
Z π
2
(1 + cos(x))2 dx
13.
Z
0
dx
sin (x) cos2 (x)
2
3.
Z
4.
sec6 (x)dx
14.
Z
dx
(sec(x) − tan(x))2
Z
sin(2x) cos(3x)dx
15.
Z
5.
Z
cos(5x) cos(7x)dx
6.
16.
Z
Z
(tan4 (x) + tan6 (x))dx
7.
Z
8.
9.
Z πs
6
0
Z p
1 + cos(x)dx
1
1
+
dx
1 − sin(x) 1 + sin(x)
18.
Z
4
tan(x)
dx
tan(2x)
17.
dx
1 + sin(x)
Z
4
cos (x) − sin (x)dx
10.
cos2 (x)
dx
2 cos2 (x) − 1
tan3 (x)dx
19.
Z
Z p
csc(x) − sin(x)dx
11.
1 + 4 cot(x)
dx
4 − cot(x)
20.
Z
Z q
2 tan2 (x) + 2 sec(x) tan(x) + 1 dx
26
sin(x) cos(x)
dx
sin4 (x) + cos4 (x)
1.16
Intermediate Trig-Substitution
In this section, we use trig-substitution in a nonstandard way. Remember that in
trig-substitution, that whole goal is to get rid of radicals or other simplification.
Using sin2 (x) Example
Z √
1−x
√ dx
1− x
[Calculus - Spivak]
Let x = sin2 θ, then dx = sin(2θ)dθ.
Z √
Z
Z
1−x
cos θ
√ dx =
sin(2θ)dθ = 2 sin θ(1 + sin θ)dθ
1 − sin θ
1− x
Now it is a much easier integral to solve. The answer will be long though, so I’d suggest
putting in bounds so that we don’t have to substitute back.
Using Double Angle Example
√
Z 1 p√
x
1−x+ 1+x
√
dx
1 − x2
0
[My High School Integration Bee Journal]
Let x = sin(2θ). Then dx = 2 cos(2θ). This substitution gets rid of
√
1 + x because 1 ± sin(2θ) = (sin θ ± cos θ)2 .
Z π
4
=
sin(2θ)
p
| sin θ − cos θ| + sin θ + cos θ2dθ
0
√
Z π √
4
3
3
2 2
=
4 2 cos 2 θ sin θdθ =
(4 − 2 4 )
5
0
27
√
1 − x or
Loophole Example
Z
ecos
−1 (x)
dx
[MIT Integration Bee 2006]
Let θ = cos−1 (x). Then cos θ = x and − sin θ = dx.
Z
−eθ sin θdθ
Identity Cancellation Example
Z 1
cos−1 (2x2 − 1)dx
0
Let x = cos θ.
Z π
2
=
−1
Z π
2
2
cos (2 cos θ − 1) sin θdθ =
0
−2θ sin θdθ
0
Note: It is true that cos−1 (cos(2θ)) = 2θ for θ ∈ [0, π2 ].
A Very Satisfying Example
Z 1
0
32
√
x − x2
r
3
x
−1
tan
dx
1−x
[UC Berkeley Integration Bee 2021]
Notice that if I test certain trig functions to plug into x, letting x = sin2 θ will
definitely simplify it better because it cancels the radicals and the arctangent.
Z π
2
=
0
32
(θ)3 sin(2θ)dθ =
sin θ cos θ
28
Z π
2
0
64θ3 dθ = π 4
1.17
Inverse Functions
Inverse Trig Identities
π
sin−1 (x) + cos−1 (x) =
2
1
π
tan−1 (x) + tan−1
=
x
2
Note: tan−1 ( x1 ) = cot−1 (x)
Inverse Trig Example
Z 1
dx
sin−1 (x)
+1
cos−1 (x)
0
Z 1
cos−1 (x)
0
π
2
=
dx =
2
π
[Harvard-MIT Math Tournament Integration Bee Mock Training]
Symmetry Example
Z π
3
cos−1 (tan(x))dx
− π3
[Silver]
Z π
3
=
−1
−1
−1
Z π
cos (tan(x)) + sin (tan(x)) − sin (tan(x))dx =
− π3
3
− π3
π
+ 0 dx
2
2
Note that sin−1 (tan(x)) is an odd function. So this integral equals to π3 .
Sum of Inverse Functions
Z f (b)
Z b
f −1 (x)dx = bf (b) − af (a)
f (x)dx +
f (a)
a
Clearly, we would rather find certain functions and bounds such that we can combine
this into one integral.
A Simple Example
Z 1
3
sin
0
πx 2
√
2
−1 3
+ sin ( x) dx
π
You can clearly see that they are inverses of each other. Hence, the answer to this
integral would be 1. But let’s be careful not to make too many integrals like this since
competitors can just easily guess.
29
A Guess-Proof Example
Z 1 √
2 x ln2 (2) + ln2 (1 + x)dx
0
[Carnegie Mellon Integration Bee 2023]
Now it’s not so obvious. Most would
guess 1, but there’s
an evil hidden √detail about
√
√
this integral. The inverse of ln2 (x + 1) is e x − 1. Notice that 2 x ln2 (2) = eln(2) x ln2 (2).
Let u = x ln2 (2) for only the first term.
Z ln2 (2) √
Z 1
u
e du +
ln2 (1 + x)dx
0
0
Wait, we’re not done. We need that -1 in the first term.
Z ln2 (2)
√
(e
u
Z 1
− 1) + 1du +
ln2 (1 + x)dx = ln2 (2) + ln2 (2)(1) − 0 = 2 ln2 (2) .
0
0
Common Substitution Example
Z 2√
1 + x3 +
√
3
x2 + 2x dx
0
Z 2√
Z 2p
3
3
=
1 + x dx +
(x + 1)2 − 1dx
0
0
Z 2√
Z 3√
3
3
=
1 + x dx +
u2 − 1du = 2(3) − 0 = 6
0
1
This is the most common way to hide inverse functions, just by u-substitutions to
manipulate the bounds.
Making a sum of inverse integral is not easy to do. You’re gonna need to play around
with functions on a graph or on WolframAlpha until you have an idea.
30
1.18
King’s/Queen’s Rule
King’s Rule Concept
Z b
Z b
f (a + b − x)dx
f (x)dx =
a
a
Z b
f (x)
u=a+b−x
dx −−−−−→
a f (x) + f (a + b − x)
Z b
f (a + b − u)
du = I
a f (u) + f (a + b − u)
Variables are just variables.
Z b
Z b
f (x) + f (a + b − x)
1dx = (b − a)
I + I = 2I =
dx =
a f (x) + f (a + b − x)
a
Hence I = b−a
.
2
A Simple Example
Z 4
3
Let u = 7 − x.
Z 4
3
√
√
x
√ dx
7−x+ x
√
7−u
√
du
√
u+ 7−u
Add it to the first integral.
√
√
Z 4
Z 4
Z 4
x
7−x
√
√
2I =
dx =
1dx
√ dx +
√
7−x+ x
x+ 7−x
3
3
3
We get 2I = 1 → I = 21 .
Logarithm Example
Z 1
ln(x + 1)
dx
2
0 ln(2 + x − x )
Z 1
Z 1
ln(x + 1)
ln(2 − x)
u=1−x
=
dx −−−−→
dx
0 ln(x + 1) + ln(2 − x)
0 ln(2 − x) + ln(x + 1)
Z 1
1
2I =
1dx ⇒ I = .
2
0
31
Flipping the Logarithm Example
x
)
1 + ln( 2−x
dx
x2 + (2 − x)2
Z 2
0
[Silver]
Let u = 2 − x, then:
1 + ln( 2−u
)
u
du
(2 − u)2 + u2
Z 2
0
Combine the first integral with our u-sub integral, the log will cancel out:
Z 2
Z 2
2
dx
π
dx =
=
.
2
2
4
0 2(x − 2x + 2)
0 (x − 1) + 1
King’s Advantage
Z 1
x
0
ex + e1−x
dx
[Silver]
If you let u = 1 − x and add the integral with our original, it’ll cancel out the
numerator:
Z 1
Z 1
x
dx
1−x
+ 1−x
dx =
x
1−x
x
x
1−x
e
+e
0 e +e
0 e +e
It is now an elementary integral we can solve!!
Trig Identity Example
Z 1
sin(x) cos(1 − x)dx
0
Z 1
Z 1
sin(x) cos(1 − x) + cos(x) sin(1 − x)dx =
2I =
0
sin(1)dx
0
Inverse Arctangent Identity Example
Z 1
−1
tan
0
Z 1
x
1−x
dx
x
1−x
−1
2I =
tan
+ tan
dx
1−x
x
0
Z
1 1π
π
I=
dx =
2 0 2
4
−1
32
Queen’s Rule Concept
If you let u = π2 − x, then
Z π
Z π
2
2
f (sin(x))dx =
0
f (cos(u))du
0
Important Trig Identities:
π
− x = cos(x)
sin
2
cos
sin(π − x) = sin(x)
π
− x = sin(x)
2
cos(π − x) = − cos(x)
Simple Queen’s Rule Example
Z π
2
0
sin3 (x)
dx
sin3 (x) + cos3 (x)
Let u = π2 − x:
Z π
2
2I =
0
cos3 (x)
sin3 (x)
+
dx =
sin3 (x) + cos3 (x) sin3 (x) + cos3 (x)
Z π
2
1dx
0
(π − x) Example
Z π
0
x sin(x)
dx
1 + cos2 (x)
Let u = π − x:
Z π
2I =
0
x sin(x)
(π − x) sin(x)
+
dx =
1 + cos2 (x)
1 + cos2 (x)
Z π
0
π sin(x)
dx
1 + cos2 (x)
Trig-Sub Example
Z ∞
0
dx
1 + x + x2 + x3
Let x = tan θ:
Z ∞
0
dx
=
(x + 1)(x2 + 1)
Z π
2
0
dθ
=
1 + tan θ
Let u = π2 − θ and you apply the same concept.
33
Z π
2
0
cos θ
dθ
cos θ + sin θ
Tangent Example
Z π
1
1
+
dx
ln(tan(x)) 1 − tan(x)
2
0
[JEE Main]
Let u = π2 − x:
Z π
2
0
Z π
2
2I =
0
1
1
+
dx
ln(cot(x)) 1 − cot(x)
1
1
1
tan(x)
−
+
−
dx
ln(tan(x)) ln(tan(x)) 1 − tan(x) 1 − tan(x)
Z π
2
π
=
1dx ⇒ I =
4
0
Note: tan( π2 − x) = cot(x),
cot( π2 − x) = tan(x)
Trig Identities Craze
Z π
2
(sin(sin2 (x)) + cos(cos2 (x)))2 dx
0
[Silver]
Z π
2
sin2 (sin2 (x)) + cos2 (cos2 (x)) + 2 sin(sin2 (x)) cos(cos2 (x))dx
0
π
Let u = 2 − x:
Z π
2
sin2 (cos2 (u)) + cos2 (sin2 (u)) + 2 sin(cos2 (u)) cos(sin2 (u))du
0
Z π
2
2I =
1 + 1 + 2 sin(sin2 (x) + cos2 (x))dx
0
I = (1 + sin(1))
34
π
2
1.19
Gaussian Integrals
Gaussian Integral Formula
√
Z ∞
−x2
e
dx =
0
π
2
Substitution Examples
Z ∞
2
2−x dx
0
Z ∞
=
0
√
Z ∞
r
u=x ln(2)
1
π
1
− ln(2)x2
−u2
e
dx −−−−−−→ p
e du =
2 ln(2)
ln(2) 0
Z ∞
2
e−x +2x dx
−∞
Z ∞
=e
√
2
e−(x+1) dx = e π
−∞
Symmetry Example
Z ∞
2
e−x (sin(x) + cos(x))2 dx
−∞
Z ∞
=
−x2
e
Z ∞
(1 + sin(2x))dx =
−∞
2
2e−x dx + 0 =
0
Disguised Example
Z ∞
0
1
=
e
Z ∞
−x2
e
0
2
x2 + 1 + e x
dx
ex2 +1 (x2 + 1)
√
1
π+ π
+ 2
dx =
x +1
2e
35
√
π
Trig Example
Z π
2
2
e− sec (x) sec2 (x)dx
0
[Stanford Math Tournament Integration Bee 2022]
1
=
e
Z π
2
− tan2 (x)
e
0
1
sec (x)dx =
e
2
Z ∞
√
−u2
e
du =
0
Integration by Parts Example
Z ∞
2
x2 e−x dx
0
2
We derive x and integrate xe−x .
√
i∞ 1 Z ∞
h x
π
−x2
−x2
+
e dx =
− e
2
2 0
4
0
Note:
x
x −x2
e
= lim x2 = 0
x→∞ 2e
x→∞ 2
lim
A Beautiful U-Substitution Integral
Z ∞
2
2
e−x 3 x− 3 dx
0
[Silver]
1
2
Let u = x 3 , then du = 13 x− 3 dx.
Z ∞
3e
0
−u2
√
3 π
du =
2
36
π
2e
1.20
Special Wildcard
Usually, the last problem is the hardest. This type of integral helps us tiebreak some
competitors. Obviously, it still must be appropriate for the qualifying exam, but it can be
anything unusual, creative, and use future techniques after this chapter. Here are examples
from other qualifying exams that past further techniques than this chapter.
1.
Z ∞
0
2.
Z ∞
5.
√
ln( x + 1)
√
dx
x x
Z ∞
0
2
e−x (5x4 − 2x6 )dx
6.
−1
3.
4.
Z
Z 2π
0
tan−1 (x)
dx
x(ln2 (x) + 1)
dx
sin (x) + cos4 (x)
3 sin(x) + 4 cos(x)
dx
4 sin(x) + 3 cos(x)
4
7.
Z ∞
0
Z 1
ln(x)(ln(x) + 1)
dx
x2x−1
0
ln(1 + x + x2 )
dx
x
You’ll soon learn how to solve these after you finish reading this whole guide lmao.
37
Chapter 2
Regular Round Topics
2.1
Modified Textbook Problems
I sometimes use calculus textbook for easy problems, skeletons, or for some
inspiration. Let’s do an example of how we modified textbook problems.
Textbook Problem 1
Z r
1−x
dx
1+x
I chose this problem because I like the idea of conjugating the radicals in a way
where it becomes a u-sub and an arcsine function.
Z
√
1−x
√
=
dx = sin−1 (x) + 1 − x2 + C
1 − x2
Can we pull this trick off with a different integral? Maybe arcsecant? But that
would be using x√x12 −1 , which is a little harder to work with. How about other conjugating
functions? I thought about √x+1−1 √x−1 =
√
√
x+1+ x−1
. Hmmm, what if:
2
√
Z √
Z
x+1+ x−1
dx
√
√
dx =
x−1
x2 − 1 − x + 1
Ooooh, this looks tricky and nice. I’ll use it!!
38
Textbook Problem 2
Z
dx
1 + ex
This is a very common integral that keeps showing up in a lot of integration bees.
Has anyone ever tried squaring it?? Let’s see how the integral is.
Z
Z
Z
dx
e−2x
u
=
dx = −
du
x
2
−x
2
(1 + e )
(1 + e )
(1 + u)2
Hmmm, what if we generalize it?
Z
Z
Z
e−nx
un−1
dx
=
dx
=
du
(1 + ex )n
(1 + e−x )n
(1 + u)n
Dang, not elementary, although this does look like something we can use in advanced competitions (beta function trick). Unfortunately, we won’t do beta functions here in this integration
bee, but we’ll keep the first idea.
Textbook Problem 3
Z
x
ee +x dx
Another easy common integral that pops up a lot in integration bee. Let’s modify
it fancier. Notice the u-sub skeleton:
Z
Z
ex x
e e dx = eu du
We can turn it into an integration by parts:
Z
Z
Z
x
2 u
ex 3x
u e du = e e dx = ee +3x dx
Ooooh, that dx form gave me an integral idea:
Z
Z
Z
1
(ex −1)(ex +1) 2x
e2x +2x−1
eu du
e
e dx = e
dx =
2e
Oooooh, we could hide the substitution even more by making the u-sub skeleton into
integration by parts:
Z
Z
1
(ex −1)(ex +1) 4x
e
e dx =
ueu du
2e
Looks good!!!
39
2.2
Sneaky Algebra Manipulation
Factoring Below 1 Degree
Z 1
0
x−4
√
dx
x−2
√
√
This could be easily solved by factoring x − 4 = ( x − 2)( x + 2).
Absorption
Z
(x2 + x)2 (x2 − x + 1)2 dx
This looks like a tedious integral, but...
Z
Z
2
2 2
2
x (x + 1) (x − x + 1) dx = x2 (x3 + 1)2 dx
Substitution
Z 4
(x − 1)(x − 2)(x − 3)(x − 4)(x − 5)dx
3
Let u = x − 3:
Z 1
Z 1
(u + 2)(u + 1)u(u − 1)(u − 2)du =
u(u2 − 1)(u2 − 4)du
0
0
Let w = u2 − 1:
1
2
Z 0
w(w − 3)dw =
−1
5
12
Reverse Absorption
Z
(x2024 + x)2022 dx
This looks disgusting, however:
Z
x2022 (x2023 + 1)2022 dx
Now it’s easier to see that this is just a u-substitution. Try this one:
Z q
3
ex + e 2 x dx
40
Exchanged Absorption
Z
Z
(x + 1)3 (2x4 + x3 )dx
Z
3 3
(x + 1) x (2x + 1)dx =
(x2 + x)3 (2x + 1)dx
Factoring Below Degree 1 Type 2
Z
Z
dx
√
(x + 1) x + 2x
dx
√ √ =
(x + 1 + 2 x) x
Z
dx
√ 2√
(1 + x) x
The way you form the integral truly matters. It can always throw off competitors;
either to the wrong direction or not.
41
2.3
Basic Forced U-Substitutions
Forced u-substitutions means to completely
perform u-sub without exchanging a
√
proper dx. For example, if I need to do u = x but I don’t have du = 2√1 x dx, then you must
solve for x by having u2 = x, then derive both sides to get 2udu = dx. After that, you can
start subbing the integral.
1.
Z
√
e
2.
10.
x
Z
dx
√
(1 + x)2
dx
11.
Z
sin(ln(x))dx
Z ln(10)
0
3.
Z 1
12.
√
(1 + x)8
Z
√
0
4.
5.
Z
13.
dx
p√
x+1
14.
Z
−1
2
Z
15.
√
3
Z
x 2x − 1 dx
7.
Z 8
0
8.
9.
Z
16.
dx
p
√
1+ 1+x
17.
√
tan ( x)dx
−1
Z
18.
3
ln (x)dx
42
dx
2021x − 1
√
3
Z
dx
x+1−1
dx
√
2 x+3+x
(sin (x)) dx
6.
√
ex ex − 1
dx
ex + 8
Z
√
3 2x+1 dx
Z ∞
ln(1 − e−x )
dx
ex − 1
ln(2)
Z
x ln5 (x)dx
Z q
√
1 − xdx
2.4
Gamma Function Integral
Factorial Integral Formula
Z ∞
n! =
xn e−x dx
0
Now that you know the formula or the definition of the gamma function, here are
some good examples that you can be creative with:
1.
7.
Z ∞
Z ∞
5 −x
x e dx
Z 1
8.
Z ∞
ln6 (x)dx
Z ∞
ex (x4 + x3 + x2 + x)dx
10.
Z ∞
−x2
x5 e
Z ∞
dx
1
0
5.
Z ∞
11.
−x
√
12.
xe
√
− x
ln(x)
x
1
0
x7 e x
1
0
Z ∞
x
e3x−e dx
Z ∞
4
e (x + 1) dx
6.
dx
−∞
−∞
4.
e x
0
9.
Z 0
x
√
0
3.
1
0
0
2.
1
e−x 3 x− 3 dx
Z π
2
dx
0
0
43
4
dx
dx
sec6 (x)e− tan(x) dx
2.5
Integration by Parts
Here are some good examples of integration by parts for regular rounds. Again, it’s
just a little trickier but also not too difficult.
1.
10.
Z
Z
xex
√ x
dx
e −1
Z
sin(ln(x))
dx
x3
Z
x2 sec−1 (x)dx
x ln(x + 1)dx
2.
11.
Z
x
e sin(x) cos(x)dx
3.
Z
4.
Z
5.
12.
ln2 (x)
dx
x3
13.
x
dx
1 + sin(x)
Z
sec(x)dx
14.
Z
Z
3
sec (x)dx
6.
7.
8.
Z
Z
Z
15.
ln(x + 1)
√
dx
x x
Z
sin(x) ln(sin(x))dx
16.
tan−1 (x)
dx
x2
Z
17.
x
Z
2
e sin (x)dx
9.
Z
x ln(x)
√
dx
x2 − 1
18.
x sin−1 (x)
√
dx
1 − x2
xex sin(x)dx
ln(1 + x4 )
dx
x3
Z
ln(tan(x)) cos(2x)dx
44
2.6
Trig-Identities Manipulation
Here, we use trig-identities to make a little trickier integrals. Here are some good
examples for this regular round.
1.
9.
Z p
1 + cos(x)dx
dx
1 + 8 cos2 (x)
10.
2.
Z
3.
sec(x)
dx
sec(x) − tan(x)
Z
Z
4.
dx
sin (x) cos2 (x)
11.
sin3 (x) + cos3 (x)dx
12.
π
π
sin x +
sin x −
dx
4
4
13.
Z
Z
6.
Z
r
2
sec (x)
2
5.
Z
14.
cos(x)
dx
cos(2x) − 1
q
p
tan(x) tan(x) tan(x) sin(x)dx
Z
Z
Z
Z π
4
2 tan(x)
dx
1 − tan2 (x)
√
dx
3 sin(x) + cos(x)
tan(x)
dx
tan(x) + cot(x)
ln(2 cos2 (x) − 1) cos(x)dx
0
7.
Z
8.
Z
15.
cos(2x)
dx
sin(x) + cos(x)
Z
sin(sin(x) cos(x)) cos(2x)dx
16.
p
cos(x) cos(2x)dx
Z
sec(x) csc(x)dx
45
2.7
Intermediate Trig-Substitution
It’s just like the qualifying exam chapter, except that the integrals are a bit trickier
and we have more time for each integral. Here are some good examples of intermediate
trig-substitution for this regular round.
1.
Z
2.
8.
x
√
dx
(1 − x) 1 − x2
Z 1
x2
2
(1 − x2 )
0
3
2
Z ∞
0
9.
dx
Z √2
esec
−1 (x)
√
x2 − 1dx
1
3.
Z 1
√
sin−1 ( 1 − x2 )dx
10.
−1
4.
Z 1
Z 1
−1
tan
0
5.
x
√
1 − x2
0
√
sin−1 ( x)
√
dx
1−x
dx
11.
Z 1
√
Z ∞p
1 + 1 + x2
dx
1 + x2
0
0
dx
√ √
(1 + x) x − x2
12.
6.
Z 1
0
7.
tan−1 (x)
√
√
dx
( 1 + x2 + x) 1 + x2
Z 4
1−x
√
dx
sin−1 (x) 1 − x2 + 1 − x2
−1
tan
q
√
x − 1 dx
1
13.
Z ∞
√
2
Z 1
dx
x2 (x2 − 1)
3
2
0
46
dx
√
(1 + x2 ) 1 − x2
2.8
Infinite Functions
Algebraic Functions
Z
dx
1 + 1+ 1 1
1+...
1
1
⇒y=
⇒ y2 + y − 1 = 0 ⇒ y =
y=
1
1+y
1 + 1+...
√
5−1
1
=
2
φ
Remember that y > 0, hence
Z
Z
dx
1
1 + 1+ 1
=
dx
φ
1+...
Geometry Series Convergence
Z 2023
0
10dx
(1 + x4 )(1 + x8 )(1 + x16 )...
[UC Berkeley Integration Bee 2023]
Z 2023
0
10(1 + x)(1 + x2 )dx
1 + x + x2 + x3 + x4 + ...
For 0 < x < 1, the geometric series applies. For 1 < x, the polynomial goes to infinity,
but since it’s the denominator, the whole integral goes to zero.
Z 1
10(1 − x4 )dx = 8
=
0
BE VERY AWARE ABOUT THE DOMAIN!!
Force U-Subbing the Infinance
Z π +1
2
sin(x − sin(x − sin(x − ...))dx
0
[MIT Integration Bee 2023 - Regular Round]
Let u = sin(x − u). Then sin−1 (u) + u = x →
Z 1
0
u+ √
√ 1
+1
1−u2
u
3
du =
2
1 − u2
47
du = dx.
Taylor Series
s r
Z
x
Z
x
1
1
1
+ 3!
+ 4!
+...
2!
3
x
q
√
4
x 5 x...dx
Z
dx =
x
P∞
1
n=2 n!
Z
dx =
xe−2 dx
Function Inside a Taylor Series
5x
e3x +
Z
e4x + e 4+...
3
2
Z
e3x e4x e5x
+
+
+ ...dx =
2!
3!
4!
Z
e
x
∞
X
enx
n=2
n!
dx
Z
dx =
x
(ee − ex − 1)ex dx
Iterative Bound
Z
R (R1... xdx)
1
xdx
xdx
1
[Cambridge University Integration Bee]
Z y
y2 − 1
⇒ y 2 − 2y − 1 = 0
2
1
√
Since y > 0, the integral equals to y = 1 + 2 .
y=
xdx ⇒ y =
48
2.9
Sneaky Symmetry
U-Substitution
Z 4
(x − 1)(x − 2)(x − 3)(x − 4)(x − 5)dx
2
Let u = x − 3:
Z 1
u(u2 − 1)(u2 − 4)du = 0
−1
Removing a Term
Z 2
1
2
ln2 (x)(sin−1 (ln(x)) + 1)
dx
x
Let u = ln(x):
Z ln(2)
2
Z ln(2)
−1
u (sin (u) + 1)du = 2
− ln(2)
u2 du
0
Super easy, yet intimidating or distracting for competitors. Here’s another good one
for you to try:
Z 1 5
x +1
(tan−1 (x))2 dx
2+1
x
−1
A Very Dangerous Type of Symmetry Trick
Z 2
x3 cos(x3 − 3x2 + 4x − 2)dx
0
[Silver]
We will never ever use this integral in UC Berkeley Integration Bee, so I suggest a
simpler integral, but for now I’m just showing you this dangerous symmetry concept.
Let u = x − 1 to purposely make the bound symmetrical.
Z 1
Z 1
3
3
(u + 1) cos(u + u)du =
2(3u2 + 1) cos(u3 + u)du = 2 sin(2)
−1
0
So what happened here is that we hid a nice polynomial u3 + u by letting u = x − 1, making it
uglier and unpredictable. The (u+1)3 was expressed and u3 +3u is an odd function cancelling
out of the integral. This leaves us our needed derivative 3u2 + 1 which is why we had u3 + u
inside cosine.
49
2.10
Sneaky Product Rule
Secant Hiding Spot
Z
3
sec (x) + sec(x) tan2 (x) dx
This will definitely scare the hell out of speed integrators, especially beginner calculus
mathletes. Even though each term is famous for having such an ugly answer and a terrible
integration by parts process, adding both of these integrals gives a very sneaky product rule
function.
Z
sec(x) sec2 (x) + (sec(x) tan(x)) tan(x)dx
And now you see that the answer is just sec(x) tan(x) + C .
Exponential Product Rule
Z
Z
=
e
x
ex (2x + 1)
√
dx
x
√
1
2 x+ √
x
√
= 2 xex + C
Most beginners may not be able to see this. They’ll probably just do integration by
parts. You can also trick them with this as well:
Z
ex (sin(x) − cos(x))dx
They’ll accidentally think its not product rule because of the minus sign.
Hiding the Exponential Product Rule
Z
sin(ln(x)) + cos(ln(x))dx
Let u = ln(x):
Z
eu (sin(u) + cos(u))du
50
xs and Logarithms
Z
ln(ln(x)) +
Z
1 · ln(ln(x)) + x
=
1
x ln(x)
1
dx
ln(x)
dx = x ln(ln(x)) + C
Radical Product Rule
Z r
ln(x)
1
+p
dx
x
x ln(x)
Z p
p
√
1
1
= 2 x ln |x| + C
=
ln(x) · √ + x · p
x
x ln(x)
Multi-Product Rule
Z 2π
cos(x)(x3 + 6x)dx
0
[Harvard-MIT Math Tournament Integration Bee Mock Training]
Z 2π
=
cos(x)(x3 + 6x) + sin(x)(3x2 − 3x2 )dx
0
Z 2π
=
(x3 cos(x) + 3x2 sin(x)) + (6x cos(x) − 3x2 sin(x))dx
0
2π
= x3 sin(x) + 3x2 cos(x) 0 = 12π 2
An Imposter Integral
Z ∞
2
e−x (5x4 − 2x6 )dx
−1
[Silver]
No, this is not a Gaussian integral.
Z ∞
4 −x2
5
−x2
+ x −2xe
dx
=
5x e
−1
h
i∞
1
5 −x2
=
= xe
e
−1
51
Chapter 3
Intermediate Topics
3.1
Power Substitution
Getting Rid of Radicals
√
Z p
1− x
dx
x
Let u =
p
√
1 − x, then (1 − u2 )2 = x → −4(1 − u2 )u = dx.
Z
Z
−4u2 (1 − u2 )
u2
du
=
4
du
(1 − u2 )2
u2 − 1
Power Subbing
Z 1
0
dx
√
1+ 3x
√
Let u = 1 + 3 x → (u − 1)3 = x, then 3(u − 1)2 du = dx.
Z 2
Z 2
3(u − 1)2
1
du = 3
u−2+
du
u
u
1
1
Fulfilling All Radicals
Z
√
dx
√
x+ 3x
Let u6 = x so that all radicals are cancelled.
Z
Z
Z
6u5 du
6u3 du
1
=
du = 6 u2 − u + 1 −
du
3
2
u +u
u+1
1+u
52
3.2
Sneaky U-Sub Tricks
Inverse Functions
Note that W (x) is the inverse function of xex .
Z e
eW (x) dx
0
[Silver]
Let u = W (x). Then ueu = x → eu (u + 1)du = dx.
Z 1
e2u (u + 1)du
0
Trig Manipulation
sec(x) − tan(x)
p
dx
sin(x)
Z
Z
=
1 − sin(x)
p
dx =
cos(x) sin(x)
Z
cos(x)dx
p
=
(1 + sin(x)) sin(x)
Z
du
√
(1 + u) u
Trig Manipulation 2
Z π
4
0
Z π
4
=
0
tan(x)
p
dx
cos(2x)
sin(x)
p
dx =
cos(x) 2 cos2 (x) − 1
Z 1
π
du
=
2 u
4
2u2 − 1
2
√
√
Norwegian Substitution
Z ∞
0
6x
dx
4x + 9x
[Integral Kokeboken]
Z ∞
=
0
Z ∞
Z ∞
6 x
3 x
u=( 32 )x
1
du
4
2
dx =
−−−→
3 2x dx −
3
9 x
1 + (2)
ln( 2 ) 1 1 + u2
1+ 4
0
53
Blind Substitution
Z
dx
√
(2x + 1) x2 + x
Blind substitutions are very hard to come up with. Pretty
√ much there’s a hidden
manipulation inside the u-sub mechanism. Notice that if I let u = x2 + x, observe that
√
u2 = x2 + x
4u2 + 1 = 4x2 + 4x + 1
4u2 + 1 = (2x + 1).
Also,
2x + 1
du = √
dx
2 x2 + x
√
4u2 + 1
dx
du =
2u
√
2u
du = dx.
4u2 + 1
So now we have,
Z
Z
√
2u
2
1
−1
√
du
=
tan
(2
·√
du =
x2 + x) + C
4u2 + 1
u 4u2 + 1
4u2 + 1
WOOOOOOAHHH, definitely did not see that coming. It’s just a simple arctangent integral.
x
−1
But wait a minute...WolframAlpha has another answer: 2 tan
+C .
x+1
To check if both answers are valid, take the derivative of the difference of those answers
on WolframAlpha or Mathematica. If the derivative equals to zero, they are both valid. If
not, there is a domain issue. In that case, just add in bounds to the integrals, easy fix!!
Blind Substitution 2
Z 1 √ x
1+
xe dx
x
Let u =
√
xex → u2 = xex , then
2udu = ex (x + 1)dx
2u
ex (x + 1)
du
=
dx
x
u2
xe 2
1
du = 1 +
dx
u
x
54
British Substitution
Z 1
0
x−1
x+1
3
dx
x+1
[Oxford University]
2
2
2
= 1 − x+1
. Then x + 1 = 1−u
→ dx = (1−u)
Let u = x−1
2 du.
x+1
Z 0
3
u
−1
1−u
2
2
(1 − u)2
Z 0
du =
u3
du
−1 1 − u
British Substitution 2
Z 2r
1
2−x
dx
x−1
[National University of Singapore]
Let u =
q
2−x
1
. Then u2 = x−1
− 1 → u21+1 + 1 = x. Also, dx = − (u22u
du.
x−1
+1)2
Z ∞ u
0
2u
2
(u + 1)2
Z π
2
du = 2
0
sin2 θdθ =
π
2
Last integral was made by trig-subbing u = tan θ.
Needing More Power
Z ∞
1
Z ∞
=
1
dx
x x5 − 1
√
x4
1
√
dx =
5
5
5
x x −1
Z ∞
0
du
π
√ =
5
(u + 1) u
Hiding the Rational Exponent
Z
Z
=
√
x
3
2
−−−→
3
3
1 + (x 2 )2
u=x 2
√
x
dx
1 + x3
Z
3
du
2
−1
2) + C
=
tan
(x
1 + u2
3
55
Caveman Trick
Z
dx
x4 + x
This always tricks beginners. It looks like partial fractions, but...
Z
Z
dx
dx
=
3
4
x(x + 1)
x (1 + x−3 )
As simple as it looks, this is not very easy to see. Most of us aren’t comfortable with forcing
variables to have a negative exponent. But in this case, it is super helpful to do so for a
simple u-substitution.
Let u = 1 + x−3 . Then du = −3x−4 dx.
Z
1
du
1
= − ln |1 + x−3 | + C
=−
3
u
3
Again, as simple as it is, it’s not very easy to see. If you were thrown with a bunch of random
integrals, you wouldn’t be able to tell. If this is a bit too hard to see, giving more power also
works!!
Z
Z
Z
1
1
x2
1
dx
=
dx =
−
du
3
3
3
x(1 + x )
x (1 + x )
3
u u+1
=
1
1
x3 + 1
+C
ln |x3 | − ln |x3 + 1| + C = − ln
3
3
x3
Multi Logged
Z
√
ln(x) ln( 2x) ln(2x)
dx
x
[UC Berkeley Integration Bee 2022 - Tiebreaker]
√
2
)
Let u = ln(x) ln(2x). Then, du = ln(2x
dx = 2 ln(x 2x) dx.
x
LOGS EVERYWHERE
Z
x(ln2 (x) + 1)(ln(x) + 1)2 dx
[Silver]
Letting u = ln(x) is a death sentence.
Let u = x(ln2 (x) + 1). You’re welcome lol
Sneaky Note: The derivative xf (ln(x)) will give you a function full of ln(x)s.
56
3.3
Wallis’ Trick
Wallis’ Trick
For all n ∈ N,
Z π
2
sinn (x)dx =
Z π
0
2
(
cosn (x)dx =
0
π
× 12 × 34 × 65 × ... × n−1
2
n
4
6
8
n−1
2
× 5 × 7 × 9 × ... × n
3
Basic Example
Z π
2
cos4 (x)dx
0
By Wallis’ Trick,
Z π
2
cos4 (x)dx =
0
π 1 3
3π
× × =
2 2 4
16
Manipulation Example
Z π
2
sin2 (x) cos2 (x)dx
0
Z π
2
=
1
1 2
sin (2x)dx =
4
8
Z π
sin2 (u)du
0
0
π
Let w = 2 − u to turn it into Wallis’ Trick integral:
1
4
Z π
2
cos2 (u)du =
0
π
16
Helpful Wallis
Z π
2
(1 + sin(x))3 dx
0
Z π
2
(sin3 (x) + 3 sin2 (x) + 3 sin(x) + 1)dx
0
=
2 3π
π
11 5π
+
+3+ =
+
3
4
2
3
4
57
if n is even
if n is odd
3.4
Advanced Trig-Substitution
Getting Rid of the Inverse Trig Functions
r
Z 1
x
−1
dx
sin
x+1
0
Let x = tan2 θ. Then dx = 2 tan θ sec2 θdθ.
Z π
Z π
4
4
tan θ
−1
2
sin
θ · 2 tan θ sec2 θdθ
2 tan θ sec θdθ =
sec
θ
0
0
A Little Trig Manipulation
Z √2 √ 2
x −1
dx
x2 + x
1
Let x = sec θ.
Z π
4
0
sec θ tan2 θ
dθ =
sec θ(sec θ + 1)
Z π
4
(sec θ − 1)dθ = ln(1 +
0
An Intimidating Friendly Integral
Z 1
0
32
√
x − x2
3
r
x
−1
tan
dx
1−x
[UC Berkeley Integration Bee 2021]
Let x = sin2 θ to get rid of the inverse tangent.
Z π
2
0
32θ3
sin(2θ)dθ = π 4
sin θ cos θ
Another Arcsine Example
Z 1
0
Let x = tan θ.
Z π
4
−1
sin
0
1
sin−1
1 + x2
2 tan θ
sec2 θ
2x
1 + x2
dx
Z π
4
dθ =
2θdθ =
0
58
π2
16
√
π
2) −
4
An Iterating Cosine Trig-Sub
Z 2r
q
√
2 + 2 + 2 + x dx
0
This is also one of those integrals we will never put in this integration bee. I’m only
showing you this for the sake of the concept. As weird as this looks, there’s a cool ugly trick
for this. Let x = 2 cos(2θ). Note that 1 + cos(2x) = 2 cos2 (x).
Z πq
Z πr
q
p
√
4
4
2 + 2 + 2 + 2 cos(2θ)4 sin(2θ)dθ =
2 + 2 + 2 cos θ4 sin(2θ)dθ
0
0
Z πs
Z π
4
4
θ
θ
=
4 sin(2θ)dθ =
8 cos
sin(2θ)dθ
2 + 2 cos
2
4
0
0
Z π
π 4
9
7
32 4
sin
θ + sin
θ dθ =
8 − sin
4
4
63
16
0
The double angle trig-identity and the constant worked together and caused a chain reaction;
cancelling multiple radicals throughout the integral. Although, the constants are a mess.
Subbing Double Angle Cosine
Z 1
r
tan−1
0
1−x
1+x
!
dx
Let x = cos(2θ).
Z π
4
−1
tan
0
sin θ
cos θ
Z π
4
2 sin(2θ)dθ =
2θ sin(2θ)dθ =
0
1
2
Logarithm Example
Z 1
√
√
ln( 1 − x + 1 + x)dx
0
Let x = sin(2θ). Note that 1 ± sin(2x) = (sin(x) ± cos(x))2
Z π
4
Z π
4
ln(| sin θ − cos θ| + sin θ + cos θ)2 cos(2θ)dθ =
0
After integration by parts, you should get
2 ln(2 cos θ) cos(2θ)dθ
0
π 1
1
+ ln(2) −
4 2
2
59
Famous Indian Trig-Sub Monster!!!
p
√
√
Z p
3
6
x + 2 − x2 1 − x 2 − x2
√
dx
3
1 − x2
[JEE Main]
It’s one of those beautiful
integrals where it’s horrendous on the outside but very
√
nice on the inside. Let x = 2 sin θ. Note that 1 − 2 sin2 θ = cos(2θ).
√
√
√
Z p
3
2 sin θ + 2 cos θ 6 1 − 2 sin θ cos θ √
p
=
2 cos θdθ
3
cos(2θ)
Recall that 1 − sin(2θ) = (sin θ − cos θ)2 . We use cos θ − sin θ to cancel out cos(2θ).
√
√
Z
Z √
6
2
2 3 sin θ + cos θ 3 cos θ − sin θ √
p
2 cos θdθ = 2 3 cos θdθ
=
3
cos(2θ)
2
2
1
x
= 2 3 sin θ + C = 2 3 √
+ C = 26 x + C
2
Yes, that whole function was just a constant. Very wild.
Constants Assisting Us
√
Z 1
√
−1
1+x
dx
√
2+ 1−x
Let x = cos(2θ).
√
Z π
2 4 sin θ cos2 θ
2 cos θ
√
√
dθ
2 sin(2θ)dθ =
1 + sin θ
2 + 2 sin θ
0
Z π
2
4 sin θ(1 − sin θ)dθ = 4 − π
Z π
2
0
0
Another Use for Cosine
Z 1
√
−1
tan
3
2
√
2x 1 − x2
dx
2x2 − 1
Let x = cos θ. Note that 2 cos2 (x) − 1 = cos(2x). Then,
Z π
6
−1
tan
0
sin(2θ)
cos(2θ)
Z π
6
sin θdθ =
0
60
π
2θ sin θdθ = 1 − √
2 3
A Triple Angle Identity
Z √3
−1
tan
0
3x − x3
1 − 3x2
dx
[Cambridge University]
Okay...probably not use the identity for tangent, but maybe for sine or cosine.
Let x = tan θ. It is NOT well known that:
3 tan θ − tan3 θ
= tan(3θ)
1 − 3 tan2 θ
HOWEVER!!! Please realize that we are about to do tan−1 (tan(3θ)) = 3θ.
Be very careful that our integral WILL NOT BE:
Z π
3
3θ sec2 θdθ
0
√
√
Notice our top bound θ = π3 because of x = 3 → θ = tan−1 ( 3).
It is true that in general, − π2 ≤ tan−1 (x) ≤ π2 , but tan−1 (tan(3 · π3 )) = 3 π3 = π > π2 .
This is a domain problem, meaning we have to split our bounds.
Since we need 3θ = π2 →
θ = π6 , we must split our bounds at π6 .
is actually 0, which is a π difference than what we want, then
Since tan−1 tan 3 · π3
−1
tan (tan(3θ)) = 3θ − π when we go beyond the bound interval of π6 .
Hence, our true integral is
Z π
6
2
3θ sec θdθ +
0
Z π
π
(3θ − π) sec2 θdθ = √ − ln(8) .
π
3
6
3
Now without being too complicated like this evil integral, try to use the triple angle identity
for sine and cosine, and turn it into a trig-sub integral.
61
3.5
Awkward & Advanced Integration by Parts
Not Every Log is Derived
Z
ln(x) ln(1 − ln(x))dx
Integrating ln(x) gives us x ln(x) − x = x(ln(x) − 1) which will nicely cancel out with
the derivative.
Nonelementary Intimidation
Z
xex ln(1 − x)dx
This scares advanced mathletes because this looks like an impossible Taylor Series trick
due to the integral having no bounds. Integrating xex gives us xex − ex = ex (x − 1)
which will nicely cancel out with the derivative.
A Hidden Arctangent Cancellation
Z
3x2 − 1
√ tan−1 (x)dx
2x x
[Cambridge University]
Now this integral is super difficult to see. I honestly couldn’t figure out the trick to
this until WolframAlpha showed me that it’s just integration by parts.
Observe that
√
Z √
1
1
3 x
x2 + 1
3x2 − 1
3 x
√ =
√
√
−
−
⇒
dx = √
+C
2
2
2x x
2x x
2x x
x
Notice that with x2 + 1, we cancel it out nicely when we derive arctangent. Thus,
2
Z
√
3x2 − 1
x +1
−1
√ tan (x)dx =
√
tan−1 (x) − 2 x + C
2x x
x
The reason why I didn’t want to do integration by parts was because it didn’t look pretty to
perform, but I wouldn’t know.
Utilizing Trig-Manipulation
Z π
ln(1 + sin(x)) sin(x)dx
0
[Silver]
[− cos(x) ln(1 + sin(x))]π0 +
Z π
(1 − sin(x))dx = π − 2
0
62
3.6
Intermediate Partial Fractions
Two Methods Example
Z
1
(x2 + 6x + 13)(x2 + 6x + 10)
dx
The first method I will show you is dummying. Pretty much, you will treat any
nonlinear functions as linear for the sake of partial fractions. Notice that if I let α = x2 + 6x,
then
1
1
=
2
2
(x + 6x + 13)(x + 6x + 10)
(α + 13)(α + 10)
Now we can just perform basic partial fractions.
1
− 13
1
1
3
+
=
−
α + 13 α + 10
3(x2 + 6x + 10) 3(x2 + 6x + 13)
Now this is just an easier integral full of completing squares and arctangents:
Z
Z
1
1
1
1
dx
=
−
dx
(x2 + 6x + 13)(x2 + 6x + 10)
3
x2 + 6x + 10 x2 + 6x + 13
1
3
Z
1
1
1
1
−
dx = tan−1 (x + 3) − tan−1
2
2
(x + 3) + 1 (x + 3) + 4
3
6
x+3
2
+C
The second method is a bit more advanced. In this case, you utilize constants and perform
algebraic manipulation. Because we have x2 + 6x in both parenthesis, we will perform a zero
substitution with it. Watch very closely:
Z
Z
13 − 10
1
(x2 + 6x + 13) − (x2 + 6x + 10)
1
dx
=
dx
3
(x2 + 6x + 13)(x2 + 6x + 10)
3
(x2 + 6x + 13)(x2 + 6x + 10)
Z
1
1
1
=
−
dx
3
x2 + 6x + 10 x2 + 6x + 13
After that, the rest of the process is the same as the previous method.
Other problems like below, you just have to use Heaviside method (;-;).
Z
Z
dx
dx
2
3
1+x+x +x
(1 + x)2 (4 + x)
63
3.7
Basic Periodicity
Periodicity Concept
For any integers m, n,
Z 2π
sin(mx) sin(nx)dx = 0
0
Z 2π
sin(mx) cos(nx)dx = 0
0
Z 2π
cos(mx) cos(nx)dx = 0
0
Note: For the first and third case, m ̸= n otherwise it will be a square. If the answers
are going to be in terms of sine functions, then the bounds will make it equal to zero.
For cosines, you must be aware of the coefficients inside each cosines.
Consider the integral:
Z π
sin(23x) cos(19x)dx
0
By trig identities, we know that this function is in terms of sine functions because of the
trig-identity: sin(A + B) = sin(A) cos(B) + sin(B) cos(A). Coefficients are nasty, so let’s just
be general. p and q are some integer related to 23 and 19 whatever the result is:
π
Z π
sin(px) + sin(qx)
1
1
dx = − cos(px) −
cos(qx)
2
2p
2q
0
0
Uh oh, the answers are in terms of cosine, this is kind of bad because you actually have
to know what the values of p and q are. This is because cos(0) = cos(2nπ) = 1 and
cos(π) = cos((2n + 1)π) = −1 for n ∈ Z. These answers will not cancel out to zero.
If we consider the integral:
Z π
sin(23x) sin(19x)dx
0
then by trig-identities, this function is in terms of cosine functions.
π
Z π
cos(px) + cos(qx)
1
1
dx =
sin(px) +
sin(qx)
2
2p
2q
0
0
By periodicity, no matter what integers p, q are, it all equals to 0.
64
Periodicity Example
Z 2π
(cos(2020x) − 2020 sin(x))(cos(x) − sin(x))dx
0
[UC Berkeley Integration Bee 2020]
By inspection, cos(2020x) cos(x) is going to be in terms of cosines, which will give
answers of sine functions. So this term will become zero. cos(2020x) sin(x) will also become
zero. And 2020 sin(x) cos(x) also becomes zero as well. sin2 (x) will not become zero, thus:
Z 2π
Z 2π
2020 sin2 (x)dx
(cos(2020x) − 2020 sin(x))(cos(x) − sin(x))dx =
0
0
The reason why square’s don’t become zero is because they create additional constants:
sin2 (x) = 12 − 12 cos(2x). Hence,
Z 2π
Z 2π
1010 − 1010 cos(2x)dx =
=
1010 = 2020π
0
0
KEEP ON A LOOKOUT FOR SQUARES!!!!
An Invasion of Periodicity
Z 2π
(sin(x) + sin(2x) + sin(3x))2 dx
0
Obviously, I’m not gonna express this, so I’ll take advantage of periodicity. Notice
that each term is going to be sin(ax) sin(bx), which becomes in terms of cosines and then
integrated into functions of sines; giving us zero due to the bounds. However, we do have
squares: sin2 (x), sin2 (2x), and sin2 (3x). Hence,
Z 2π
Z 2π
2
(sin(x) + sin(2x) + sin(3x)) dx =
sin2 (x) + sin2 (2x) + sin2 (3x)dx
0
0
Z 2π
0
1
1
3 1
− cos(2x) − cos(4x) − cos(6x)dx = 3π
2 2
2
2
Since those cosines will be integrated into sines, those also become zero.
65
3.8
Speed Bashing
We now begin the art of making speed bashing integrals. These integrals are mainly
for tiebreaking competitors. Technically, we use speed bashing in all rounds as tiebreakers.
All Standard Methods in One Integral
Z 1
√
tan−1 ( 1 − x2 )dx
0
[Cambridge University]
It is very hard to create and find a good integral that utilizes all standard methods,
but this is a good example. First, trig-substitution:
Z 1
Z π
√
2
2
tan−1 (cos θ) cos θdθ
tan ( 1 − x )dx =
0
0
−1
Second, integration by parts:
−1
Z π
π
2
2
= [sin θ tan (cos θ)]0 +
0
sin2 θ
dθ
1 + cos2 θ
Third, trigonometric manipulation:
Z π
2
=0+
0
tan2 θ
dθ =
sec2 θ + 1
Z π
2
0
sec2 θ − 1
dθ =
tan2 θ + 2
Z π
2
0
sec2 θ
sec2 θ
−
dθ
tan2 θ + 2 (tan2 θ + 2)(tan2 θ + 1)
Fourth, u-substitution into partial fractions:
Z ∞
Z ∞
1
1
u2
=
−
du =
du
u2 + 2 (u2 + 2)(u2 + 1)
(u2 + 2)(u2 + 1)
0
0
Z ∞
0
Z ∞
−2
−1
2
1
−1
1
+
du
=
−
du
u2 + 2 u2 + 1
u2 + 2 u2 + 1
0
Fifth, improper integral evaluation:
∞
√
u
π √
−1
−1
√
=
2 tan
− tan (u)
= ( 2 − 1)
2
2
0
Can you find a faster solution for this integral???
66
Sprinting Example
Z
x5 ex dx
By utilizing pattern recognition,
= ex (x5 − 5x4 + 20x3 − 60x2 + 120x − 120) + C
Symmetry Example
Z 1
x+
3
√
1 − x2 dx
−1
By Symmetry and letting x = sin θ,
Z π
√
√
2
3π
3
2
2
3x 1 − x + ( 1 − x ) dx =
=2
6 sin2 θ cos2 θ + 2 cos4 θdθ = 1 +
8
0
0
Z 1
2
The following integral can be solved using Wallis’ Trick. Psssh...factor cos2 θ! ;)
Integration by Parts Example
Z
x(ex + sin(x) + 1)dx
The x can just be integrated. Instead of doing IBP separately, you could just integrate
(ex + sin(x)) together to not waste time.
Trig-Sprinting
Z
Z
=
sin3 (x) + cos3 (x) + tan3 (x)dx
sin(x)(1 − cos2 (x)) + cos(x)(1 − sin2 (x)) + tan(x) sec2 (x) − tan(x)dx
Algebra Sprinting
Z
Z
1
2
x +1
1
2
x +1
1
1
2
+ + 1 + x + x dx
x2 x
Z
1
1
1
1
1
2
+1 + x+
+ x dx =
+ +1− 2
dx
2
2
x
x
x
x
x +1
67
Here are some speed bashing integral examples I’ve made over the past years:
1.
8.
s √
√
√
e e x+ x
dx
x
0
[UC Berkeley Integration Bee 2023]
Z ln2 (4)
q
Z 1q
√
√
2
x + x − 1 + x − x2 − 1dx
−1
9.
2.
Z 1
15(1 +
√
3
−x2
e
Z
x(x + 1)
dx
(x − 1)x + 1
3
(x + 1) dx
0
4.
5.
Z ∞
dx
1
x(xln(4) + 1)
[Harvard-MIT Math Tournament
Integration Bee Mock Training]
dx
11.
Z ∞
Z ∞
dx
2
2
−∞ x + πx + π
6.
Z ∞
0
q √
x dx
1
10.
Z ∞
−1
sec
x)3 dx
−1
3.
Z 4
0
12.
√
ln( x + 1)
√
dx
x x
Z ∞
0
[Stanford Math Tournament Integration
Bee 2022]
13.
dx
ex +
1 dx
ex +1
dx
p
√
1 + ex
Z 9 √ √
e 1+ x dx
0
7.
14.
Z π
sin(x) + cos(x) + tan(x)
dx
sec2 (x)
− π2
2
68
Z
tan(2x)
dx
tan2 (x)
Chapter 4
A Little Advanced Topics
4.1
Undertow Technique
Beginner Example
Z
2x3 − 1
dx
x4 + x
[MIT Integration Bee 2006]
This confuses a lot of beginners because the derivatives don’t match and it looks
awkward. The worst thing most students do is performing partial fractions. But it does
look very sus though because the numerator does look like the derivative of the denominator.
Watch happens when we keep dividing top and bottom by x:
Z
Z
Z
2x2 − x1
2x − x12
2x3 − 1
dx =
dx =
dx
x4 + x
x3 + 1
x2 + x1
This is very difficult to see if you haven’t been speed integrating a lot, but it is common to
know for speed integration that:
1
d 1
= − 2.
dx x
x
A good thing to keep in mind is that reciprocal power rule alternates the sign. In this case,
1
we have our answer: ln x2 +
+C
x
Here’s a very difficult version:
Z
4x5 − 1
dx
(x5 + x + 1)2
[UC Berkeley Integration Bee 2020]
69
Nonlinear Functions Example
Z
sin(x) + cos(x)
dx
ex + cos(x)
[JEE Main]
Z
=
sin(x) + cos(x)
dx
·
x
e
1 + cos(x)
x
e
−x
Let u = 1 + e
cos(x).
A Sneaky Exponential Example
Z
x2 + e x
dx
x2 + xex
[Silver]
Z
x2 e−x + 1
dx =
x2 e−x + x
Z
e−x + x12
dx
e−x + x1
This is super difficult to see, but let u = e−x + x1 .
Logarithm Example
Z
1 − 2 ln(x)
dx
x(x2 + ln(x))
This integral actually requires experimentation in order to figure out the answer to
this. Let’s factor an x out of the denominator.
Z
1 − 2 ln(x)
dx
x2 (x + ln(x)
)
x
If we let u = x + ln(x)
, then du = 1 + 1−ln(x)
dx. This substitution does not work unfortunately,
x
x2
so we must divide x again.
Z
1 − 2 ln(x)
dx
x3 (1 + ln(x)
)
2
x
If we let u = 1 + ln(x)
, then du = x−2xx4ln(x) dx = 1−2xln(x)
dx. Jackpot!!!
3
x2
Z
du
ln |x|
= ln 1 + 2 + C
u
x
70
4.2
Square-Square Technique
Square-Square Concept
2
1
1
x+
= x2 + 2 + 2
x
x
2
1
1
= x2 + 2 − 2
x−
x
x
2
2
1
1
1
⇒ x+
−2= x−
+ 2 = x2 + 2
x
x
x
1
1
d
x+
=1− 2
dx
x
x
d
1
1
x−
=1+ 2
dx
x
x
Beginner Example
Z
x2 + 1
dx
x 4 − x2 + 1
Everytime you see a bunch of squares, a good thing to come in mind is to utilize
the square-square technique. Divide everything by x2 . This will make us complete a square
using the square-square concept.
Z
1 + x12
dx
x2 − 1 + x12
To determine which way to complete the square, we follow our du or the derivative on the
numerator. Because we have 1 + x12 , we want to do u = x − x1 , so we complete the square in
that form.
Z
Z
1 + x12
1 + x12
dx =
dx
(x2 + x12 − 2) + 2 − 1
(x − x1 )2 + 1
1
−1
1
Hence, performing u = x − x gives us tan
x−
+C .
x
71
Trig-Sub Example
x2 − 1
dx
√
·
x2 + 1
1 + x4
Z
[Calculus - Spivak]
Z
Z
1 − x12
(x2 − 1)
q
q
dx =
dx
x(x + x1 )x x2 + x12
(x + x1 ) x2 + x12
Z
Z
1 − x12
du
q
√
dx =
u u2 − 2
(x + x1 ) (x + x1 )2 − 2
Zero Substitution Example
Z ∞
0
Z
1 ∞ 1 + 1 + x12 − x12
dx =
2 0
x2 + x12 + 2
x2 + x12 + 2
0
0
∞
Z
1 + x12
x − x1
1 ∞ 1 − x12
1
1
π
1
−1
dx =
tan
=
−
1 2 +
1 2
1 +
2 0 (x + x )
2
2
2
4
(x − x ) + 4
x+ x
0
Z ∞
dx
1
1 =
2
2
x + 2 + x2
Z ∞
dx
(x + x1 )2
2
Trig Example
Z π
2
0
dx
sin (x) + cos4 (x)
4
Let u = tan(x).
Z π
Z ∞
sec2 (x)(1 + tan2 (x))
1 + u2
dx
=
du
1 + tan4 (x)
1 + u4
0
0
Z ∞
Z ∞
1 + u12
1 + u12
π
du = √
1 du =
1 2
2
u + u2
(u − u ) + 2
2
0
0
2
72
4.3
1 + sin(2x) Trick
1 + sin(2x) Concept
Very Similar to the Square-Square Technique
1 + sin(2x) = (sin(x) + cos(x))2
1 − sin(2x) = (sin(x) − cos(x))2
Beginner Example
cos(x) − sin(x)
p
dx
1 + 1 + sin(2x)
Z
Z
cos(x) − sin(x)
dx = ln |1 + sin(x) + cos(x)| + C
1 + sin(x) + cos(x)
Trig-Sub Example
Z 1
−1
Let x = sin θ:
√
√
1 − x2 − x
√
dx
1 − x2 (1 + x 1 − x2 )
Z π
cos θ − sin θ
dθ =
− π2 1 + sin θ cos θ
Z π
2(cos θ − sin θ)
dθ
− π2 2 + 2 sin θ cos θ
Z π
Z 1
2
2(cos θ − sin θ)
du
dθ = 2
= π
2
2
− π2 1 + (sin θ + cos θ)
−1 u + 1
2
2
Completing the Square
Z
Z
cos(x) − sin(x)
p
dx
sin(2x)
cos(x) − sin(x)
p
dx =
(sin(x) + cos(x))2 − 1
73
Z
√
du
u2 − 1
Trig Manipulation
Z p
p
tan(x) + cot(x)dx
Z
√ Z
sin(x) + cos(x)
sin(x) + cos(x)
p
dx = 2 p
dx
sin(x) cos(x)
1 − (sin(x) − cos(x))2
√ Z
√
du
= 2 √
= 2 sin−1 (sin(x) − cos(x)) + C
1 − u2
Zero Substitution Example
Z
dx
sec(x) + sin(x)
[JEE Main]
Z
Z
Z
2 cos(x)
cos(x) + cos(x) + sin(x) − sin(x)
cos(x)
dx =
dx =
dx
1 + sin(x) cos(x)
2 + sin(2x)
2 + sin(2x)
Z
cos(x) − sin(x)
sin(x) + cos(x)
=
+
dx
2
1 + (sin(x) + cos(x))
3 − (sin(x) − cos(x))2
Z
Z
du
dw
=
+
1 + u2
3 − w2
A Satisfying Example
p
Z π p
4
cot(x) − tan(x)
dx
1 + sin(2x)
0
Z π
4
0
cos(x) − sin(x)
p
dx
sin(x) cos(x)(sin(x) + cos(x))2
2
Let u = sin(x) + cos(x). Then u 2−1 = sin(x) cos(x).
Z √2
1
√
Z π√
4
2
√
du =
2 cos θdθ = 1
u2 u2 − 1
0
74
4.4
Advanced King’s/Queen’s Rule
Disguised Example
Z 2√
x2 − x + 1 −
√
x2 − 3x + 3dx
0
Let u = 2 − x:
Z 2p
p
(2 − u)2 − (2 − u) + 1 − (2 − u)2 − 3(2 − u) + 3du
0
=
Z 2√
u2 − 3u + 3 −
√
u2 − u + 1du = −I
0
If the original integral I = −I, then I = 0 .
Exponential Example
Z 1
0
x
dx
x + (x − x2 )x
[Silver]
Z 1
=
0
x
dx =
x + xx (1 − x)x
Z 1
0
x1−x
dx
x1−x + (1 − x)x
Taking Advantage of King’s Rule
Z 1
x
2
2
ln(x − x ) ln
+ ln(x) ln(1 − x) dx
1−x
0
[Silver]
After some algebra simplification:
Z 1
Z 1
3
3
ln (x) + ln (1 − x)dx = 2
ln3 (x)dx = −12
0
0
The last integral can be solved using Gamma function.
75
Trig Example
Z 1
0
tan(x)
dx
1 − tan(x) tan(1 − x)
[Harvard-MIT Math Tournament Integration Bee Mock Training]
Let u = 1 − x:
Z 1
=
0
tan(1 − x)
dx
1 − tan(x) tan(1 − x)
Z 1
tan(x + 1 − x)dx ⇒ I =
2I =
0
tan(1)
2
A Crazy Example!!
x
4
Let f (x) = 2+4
x . Then compute,
Z 3
4
f (f (x))dx
1
4
[Harvard-MIT Math Tournament Integration Bee Mock Training]
x
x
4
2
This is actually very hard to see. Notice that f (x) = 2+4
x = 21−x +2x .
Because of this formation, it is sneaky to see that f (x) + f (1 − x) = 1.
Also notice that f ( 41 + 34 − x) = f (1 − x). So we can definitely use King’s Rule!!
By u = 1 − x,
Z
Z
b
b
f (f (1 − x))dx
f (f (x))dx =
a
a
Since f (1 − x) = 1 − f (x),
Z b
Z b
f (f (1 − x))dx =
a
Z b
f (1 − f (x))dx =
a
So now
1 − f (f (x))dx
a
Z 3
4
2I =
1dx ⇒ I =
1
4
1
.
4
MINDBLOWING!!! I don’t think I can ever make another integral like this though ;-;
Definitely one of the most unique King’s Rule integral I’ve ever made.
76
A Very Famous Queen’s Rule
Z π
2
ln(sin(x))dx
0
Like most beginners, you can easily mess up the process if you’re not careful enough.
Usually, I’d just memorize the answer to this integral. Let u = π2 − x. Note that x and u are
just variables.
Z π
Z π
2
2
ln(cos(x))dx ⇒ 2I =
ln(sin(x) cos(x))dx
0
Z π
0
Z π
1 2
1
1 2
I=
ln
sin(2x) dx =
− ln(2) + ln(sin(2x))dx
2 0
2
2 0
Z
π ln(2) 1 π
=−
+
ln(sin(u))du
4
4 0
Observe closely that
Z π
Z π
ln(sin(u))du =
ln(sin(u))du +
0
So we have
Z π
2
ln(sin(u))du
π
2
0
1
π ln(2) 1
+ I+
I=−
4
4
4
Z π
ln(sin(x))dx
π
2
Then u = π2 − x gives,
ln(sin(x))dx =
π
2
Z π
Z 0
Z π
2
ln(cos(u))du =
− π2
ln(cos(u))du = I
0
Hence,
π ln(2) 1
+ I
4
2
1
π ln(2)
I=−
2
4
π
I = − ln(2)
2
I=−
That was a lot to take in. As you can see, you can easily get lost in the process. And a lot of
integrals keep using this famous integral in different forms. Because of that, I just remember
the answer of this integral.
77
Trig-Manipulation
Z π
2
0
p
cos(x)
p
p
dx
( sin(x) + cos(x))5
[JEE Main]
Let u = π2 − x:
p
Z π
sin(x)
dx
1 2
p
p
p
p
dx =
2 0 ( sin(x) + cos(x))4
( sin(x) + cos(x))5
Z π
2
=
0
1
2
Z π
2
0
Z ∞
sec2 (x)
1
p
dx =
4
2
( tan(x) + 1)
0
du
1
√
=
4
6
( u + 1)
Absolute Value Example
Z π
2
π
e|x− 4 | sin2 (x)dx
0
[Silver]
Let u = π2 − x:
Z π
2
e
=
|x− π4 |
0
1
cos (x)dx =
2
2
Z π
2
π
π
e|x− 4 | dx = e 4 − 1
0
Just King’s Rule
Z π
6
0
x
dx
cos(x) cos( π6 − x)
[Cambridge University Integration Bee]
Let u = π6 − x:
π
12
Z π
6
0
dx
π
=
π
cos(x) cos( 6 − x)
6
π
6
Z π
6
0
Z π
6
0
dx
√
cos(x)( 3 cos(x) + sin(x))
sec2 (x)
π
√
dx =
ln
6
( 3 + tan(x))
78
4
3
The Evil Queen’s Rule
Z π
4
ln(1 + tan(x))dx
0
This is super hard to see, yet the trig-manipulation is so satisfying!!!
Let u = π4 − x. Note that:
tan
Then,
Z π
4
0
1 − tan(x)
−x =
4
1 + tan(x)
π
Z π 4
1 − tan(x)
2
ln 1 +
dx =
ln
dx
1 + tan(x)
1 + tan(x)
0
Z π
4
π
2I =
ln(2)dx ⇒ I =
ln(2)
8
0
Who would’ve thought that algebra manipulation would cause the logarithms to cancel out
so nicely.
Trigonometric Algebra
Z π
2
0
sin3 (x)
dx
1 − sin(x) cos(x)
[Silver]
Z π
2
0
Z π
2 (s + c)(s2 − sc + c2 )
sin3 (x) + cos3 (x)
dx =
2(1 − sin(x) cos(x))
2(1 − sc)
0
Z π
2 sin(x) + cos(x)
dx = 1 .
=
2
0
79
4.5
Weierstrass Substitution
Weierstrass Concept
Let u = tan x2 , then:
2
du
1 + u2
2u
sin(x) =
1 + u2
1 − u2
cos(x) =
1 + u2
dx =
The purpose of this technique is to turn ugly trig integrals into rational polynomials.
Please use this method sparringly or you’ll end up making the integral worse.
Beginner Example
Z 2π
0
dx
2 + cos(x)
) will go beyond the
Notice we can’t exactly do this substitution because tan( 2π
2
domain. However, because this function is periodic at 2π, we can actually translate the
function since the bounds add up to 2π:
Z π
Z 2π
dx
dx
=
2 + cos(x)
−π 2 + cos(x)
0
Let u = tan( x2 ):
Z ∞
1
2du
=
·
2
1−u2
−∞ 2 + ( 1+u2 ) 1 + u
Z ∞
2du
=
2
2
−∞ 2 + 2u + 1 − u
Z ∞
2du
2π
= √
2
3
−∞ u + 3
BE VERY CAREFUL ABOUT THE DOMAIN. Sometimes, tangent functions are dangerous
to mess around with if you’re not careful enough.
A Log Example
Z π
2
0
Let u = tan( x2 ):
Z 1
du
0
2u
1−u2
1 + ( 1+u
2 ) + ( 1+u2 )
=
Z 1
=
0
dx
1 + sin(x) + cos(x)
2du
=
2
1 + u + 2u + 1 − u2
80
Z 1
0
·
2
1 + u2
du
= ln(2)
u+1
4.6
Reverse Quotient Rule
Reverse quotient rule works the same way as reverse product rule. However, it’s so
much harder to solve than reverse product rules. Things can get super ugly and sometimes,
you have to find the right function that fits the quotient rule.
A Simple Example
Z
xe2x
dx
(2x + 1)2
We know that by quotient rule, we can assume g(x) = (2x + 1). Then,
f ′ (x)(2x + 1) − f (x)(2) = xe2x
What if f (x) = e2x ? Then,
2e2x (2x + 1) − 2e2x = 4xe2x
Aha, we just need that 4. Hence, our answer for the integral is
e2x
+C .
4(2x + 1)
As you can see, it’s not an easy process as the reverse product rule. You actually have
to guess and check for functions that best fits the function in the integral.
Logarithm Example
Z
ln(x)
dx
(1 + ln(x))2
Let g(x) = 1 + ln(x), then f ′ (x)(1 + ln(x)) − f (x) x1 = ln(x).
If f (x) = x, then (1 + ln(x)) − 1 = ln(x), which matches the numerator.
Trig Example
Z
sin2 (x)
dx
(x cos(x) − sin(x))2
Assume g(x) = x cos(x) − sin(x), then
f ′ (x)(x cos(x) − sin(x)) − f (x)(−x sin(x)) = sin2 (x)
The first thing I saw was wanting to cancel out f ′ (x)x cos(x) with f (x)x sin(x).
This works by letting f (x) = cos(x).
cos(x)
This also gives sin2 (x), so our answer is
+C .
x cos(x) − sin(x)
81
A Sinister Example
Z tan−1 (x)
x − tan−1 (x)
2
dx
[JEE Main]
This is probably the most evil integral to ever give in an integration bee.
2
2
x
dx =
− 1 dx
x − tan−1 (x)
Z 2
Z
x − 2x(x − tan−1 (x))
2x tan−1 (x) − x2
=
+
1dx
=
+ 1dx
(x − tan−1 (x))2
(x − tan−1 (x))2
Z tan−1 (x)
x − tan−1 (x)
Z Let g(x) = x − tan−1 (x). Then
f (x)(x − tan (x)) − f (x) 1 −
′
−1
1
2
x +1
= 2x tan−1 (x) − x2
I need that x2 + 1 to cancel out. So let f (x) = x2 + 1.
2x(x − tan−1 (x)) − x2 = x2 − 2x tan−1 (x)
How evil, trying to leave out a negative.
So our answer is (DON’T FORGET THE x!!!) x −
x2 + 1
+C
x − tan−1 (x)
When I first tried to solve this integral, I literally tried doing quotient rule with the tan−1 (x)
on top. It was so terrible, I didn’t know I was suppose to perform zero-substitution first and
then reverse quotient rule.
82
Chapter 5
Advanced Topics
5.1
Semi-Integration by Parts
Beginner Example
Z
2
ex (2x2 + 1)dx
Sometimes it’s super difficult to see the reverse product rule. So it’s better not to
waste time trying to think which function is which. Instead, whenever we see each term being
nonelementary, we perform semi-integration by parts. Instead of integrating by parts with
2
the whole integral, we will only do it to the first term 2x2 ex .
Z
Z
Z
Z
2
2 x2
x2
x2
x2
2x e dx + e dx = xe − e dx + ex dx
2
The nonelementary integrals satisfyingly cancels out, so that gives us our answer xex + C .
A Double Example
Z
esin(x) ·
x cos3 (x) − sin(x)
dx
cos2 (x)
[Calculus - Spivak]
Let’s simplify the integral further:
Z
esin(x) (x cos(x) − sec(x) tan(x))dx
We will do IBP on each terms differently.
Z
Z
sin(x)
sin(x)
sin(x)
xe
− e
dx − e
sec(x) + esin(x) dx
The nonelementary integrals cancels out, giving us esin(x) (x − sec(x)) + C .
83
5.2
Advanced Trig Integrals
A Little Scary Example
Z
dx
9 sin (x) + 4 cos2 (x)
2
Z
sec2 (x)
dx =
9 tan2 (x) + 4
A Constant Changes Everything
Z
√
Z
du
9u2 + 4
dx
3 sin(x) + cos(x)
[JEE Main]
1
2
Z
dx
1
√
=
3
1
2
sin(x) + cos(x)
2
2
Z
π
sec x −
dx
3
Linear Algebra???
Z
sin(x) + 2 cos(x)
dx
3 sin(x) + 4 cos(x)
[JEE Main]
We want to turn this function into two different rational functions. One function
where the numerator is the same as the denominator and the other function where the numerator is the derivative of the denominator. In that way, we have two easy integrals: one
term just being a constant and the other being a simple u-sub logarithm. But in order to do
that, we must find the appropriate coefficients:
α(3 sin(x) + 4 cos(x)) + β(3 cos(x) − 4 sin(x)) = sin(x) + 2 cos(x)
This gives us a system of equations by separating in terms of sines and cosines:
11
2
3α − 4β = 1
⇒α= , β=
4α + 3β = 2
25
25
Now that we have our coefficients, our integral is now easier:
Z
11
2 3 cos(x) − 4 sin(x)
11
2
+
dx =
x+
ln |3 sin(x) + 4 cos(x)| + C
25 25 3 sin(x) + 4 cos(x)
25
25
84
A Difficult Trig-Manipulation
Z
dx
sin(x + 2) sin(x + 3)
[JEE Main]
1
sin(1)
1
=
sin(1)
Z
Z
sin((x + 3) − (x + 2))
dx
sin(x + 2) sin(x + 3)
sin(x + 3) cos(x + 2) − cos(x + 3) sin(x + 2)
dx
sin(x + 2) sin(x + 3)
Z
1
=
cot(x + 2) − cot(x + 3)dx
sin(1)
Double Angle Identities Chaos
Z
cos8 (x) − sin8 (x)dx
Let c = cos(x) and s = sin(x). Then,
c8 − s8 = (c4 + s4 )(c4 − s4 )
= (c4 + s4 ) cos(2x)
= (c2 (1 − s2 ) + s2 (1 − c2 )) cos(2x)
= (1 − 2s2 c2 ) cos(2x)
1 2
= 1 − sin (2x) cos(2x).
2
So now our integral is just
Z 1 2
1 − sin (2x) cos(2x)dx
2
This simplified a lot nicer than I expected.
85
A Chain of Tangents
Z
tan(x) tan(2x) tan(3x)dx
[JEE Main]
This is also super difficult to see, especially when you’re not comfortable with tangent
trig-identities. Observe that,
tan(x) + tan(2x)
1 − tan(x) tan(2x)
tan(3x)(1 − tan(x) tan(2x)) = tan(x) + tan(2x)
tan(3x) − tan(2x) − tan(x) = tan(x) tan(2x) tan(3x).
tan(3x) =
So our integral is
Z
tan(3x) − tan(2x) − tan(x)dx
Again!! Waaay nicer than I expected. Although, manipulating trig-identities like this is
super difficult to get comfortable with. I would consider sticking with easier identities for this
integration bee.
A Chain of Sines
Z π
2
sin(x) sin(2x) sin(3x)dx
0
[MIT Integration Bee 2010]
Note that sin(x) sin(3x) = 12 (cos(2x) − cos(4x)). Then,
1
2
Z π
2
0
1
sin(2x)(cos(2x) − cos(4x))dx =
2
Z π
2
sin(2x)(cos(2x) − 2 cos2 (2x) + 1)dx
0
Let u = cos(2x) and apply symmetry:
1
=
4
Z 1
1
(1 + u − 2u )du =
2
−1
2
86
Z 1
0
(1 − 2u2 )du =
1
6
Sneakiest Trinomial Conjugation
Z π
4
0
sin(x)
dx
sec(x) + tan(x) + 1
[JEE Main]
It is NOT well known that
(sec(x) + tan(x) + 1)(1 + tan(x) − sec(x)) = 2 tan(x)
We will use this to conjugate this nasty integral:
Z π
4
0
Z π
4
=
0
1 + tan(x) − sec(x)
sin(x)
·
dx
sec(x) + tan(x) + 1 1 + tan(x) − sec(x)
1
sin(x) + sin(x) tan(x) − tan(x)
dx =
2 tan(x)
2
Z π
4
cos(x) + sin(x) − 1dx
0
Look how beautiful that turned out. Just one sneaky trig-manipulation literally turned a
monster integral into a very easy integral. Obviously, we will not use this integral in our
integration bee. Consider making an easier function or a trig-identity that is more known and
easier to see.
Just Simplification
Z
ln
Z
ln
2 sin(x) cos(x)
2 cos2 (x)
sin(2x)
1 + cos(2x)
sec2 (x)dx
Z
2
sec (x)dx =
87
ln(tan(x)) sec2 (x)dx
5.3
Advanced Gaussian Integral
Gaussian Integral in Log Form
Z 1s 1
ln
dx
x
0
Let u =
q
2
2
ln( x1 ). Then e−u = x → −2ue−u du = dx.
Z ∞
2 −u2
2u e
0
√
π
du =
2
A Very Cool Gaussian Log
Z ∞
2
e− ln (x) dx
0
[UC Berkeley Integration Bee 2023]
Let u = ln(x):
Z ∞
−u2 +u
e
Z ∞
du =
−∞
−(u2 −u+ 41 )+ 14
e
du = e
1
4
Z ∞
−∞
−(u− 12 )2
e
du =
−∞
More Log Integrals
Z ∞p
1 + ln(x)
dx
1
x2
e
[Silver]
Let u =
p
2
1 + ln(x), then x = eu −1 .
Z ∞
0
2
u · 2ueu −1
du =
e2u2 −2
Z ∞
2 −u2 +1
2u e
0
√
e π
du =
2
This Converges???
Z ∞
1
1 − e− x2 dx
0
Let u = x1 :
Z ∞
0
"
#∞
Z ∞
2
2
√
1 − e−u
1 − e−u
2
du =
+2
e−u du = π
2
u
u
0
0
88
q
√
π e
Radical Form of Gaussian Integral
Z ∞
ex
1
dx
√
x−1
[UC Berkeley Integration Bee 2023]
Let u =
√
x − 1:
Z ∞
0
2udu
2
2 +1 du =
u
e
e
u
Z ∞
e
−u2
0
√
π
du =
e
A Beautiful Gaussian Integral
Z ∞
0
Z ∞
=
ln( √1x ) ln(x)
e
1
√
x
ln(x)
dx
Z ∞
dx =
0
1
2
e− 2 ln (x) dx =
√
2πe
0
A Crazy Lambert W Function Integral
Z ∞
W
0
1
x2
dx
Recall that W (x) is the inverse function of xex . Let u = W
ueu =
1
x2
1
x2
2
eu (1 + u)du = − 3 dx
x
u+1
− √ u = dx
2u ue
That was a lot of manipulation, so now we have
Z ∞
Z ∞
Z ∞ √u + √1
u+1
2w + 1
u
√
du =
dw
√ u du =
u
2e 2
2 ue 2
2wew
0
0
0
Z ∞√
√
2
2(2t2 + 1)e−t dt = 2π
0
89
. Then,
5.4
Dirichlet Integral
Dirichlet Integral Formula
Z ∞
sin(x)
π
dx =
x
2
0
Substitution Example
Z ∞
0
Let u = x3 :
Z ∞
0
sin(x3 )
dx
x
sin(x3 )x2
dx =
x3
Z ∞
0
sin(u)
π
du =
3u
6
Trig-Identity Example
Z ∞
sin(5x) cos(3x)
dx
x
0
Z ∞
=
0
sin(8x) + sin(2x)
π
dx =
2x
2
IBP Example
Z ∞
0
1 − cos(x)
dx
x2
Derive 1 − cos(x) and integrate x12 :
∞ Z ∞
cos(x) − 1
sin(x)
π
=
+
dx =
x
x
2
0
0
Exponential Example
Z ∞
sin(ex )dx
−∞
Let u = ex :
Z ∞
0
sin(u)
π
du =
u
2
90
Trig U-sub Example
Z π
2
sin(tan(x))
dx
sin(2x)
Z ∞
sin(u)
π
du =
2u
4
0
Let u = tan(x):
0
IBP Example 2
Z ∞
0
sin3 (x)
dx
x3
Derive sin3 (x) and x13 twice.
∞ Z ∞
3 sin2 (x) cos(x) sin3 (x)
3 sin(2x) 3 sin3 (x)
=
−
+
−
dx
2x
2x2 0
2x
2x
0
Z ∞
sin(2x)
3 sin(x)( 21 − 12 cos(2x))
−
dx
=0+
x
2
x
0
Z
3π 3π 3 ∞ sin(3x) − sin(4x)
3π
=
−
+
dx =
4
8
4 0
x
8
3
2
Intimidating Example
Z ∞
0
Z ∞
=
0
(x − 1) sin(x) + x cos(x)
dx
x2
sin(x) x cos(x) − sin(x)
π
+
dx =
−1
2
x
x
2
91
Chapter 6
Very Advanced Topics
6.1
Inverse Jailbreaking
Inverse Jailbreaking Concept
We learned that
Z b
Z f (b)
f (x)dx +
a
f −1 (x)dx = bf (b) − af (a)
f (a)
Let’s say we come across a disgusting definite integral. Is the inverse of the function
easier to integrate??? If so, we can perform:
Z f (b)
Z b
f (x)dx = bf (b) − af (a) −
f −1 (x)dx
f (a)
a
Beginner Example
Z 2
√
csc−1 ( x)dx
1
[Caltech Homework]
I don’t wanna
inverse cosecants, so let’s integrate the inverse instead.
√ deal with −1
−1
If y = csc ( x), then f (x) = csc2 (x).√ If the bounds were from 1 to 2, then our new
bounds are from csc−1 (1) = π2 to csc−1 ( 2) = π4 .
Z 2
π π
csc (x)dx = 2 · − −
4
2
1
−1
92
Z π
4
π
2
csc2 (x)dx = 1
Thank you symmetry!!
Z 8
√
tan−1 ( 3 x − 1)dx
0
[Silver]
The inverse function of that is (tan(x) + 1)3 . Our new integral is now:
=8
π 4
Z π
4
−0−
Z π
3
4
(tan(x) + 1) dx = 2π − 2
− π4
3 tan2 (x) + 1 dx
0
No leftovers.
Z 1
r
tan−1
3
0
1−x
x
!
dx
[Silver]
The inverse function for this integral is 1+tan1 3 (x) . Notice that
r
Z 1
3 1 − x
x
−1
tan
0
!
Z 0
π dx
=
1(0)
−
0
3
π 1 + tan (x)
2
2
dx +
This means that
Z 1
r
−1
tan
0
3
1−x
x
!
Z π
2
dx =
0
π
dx
=
3
1 + tan (x)
4
A Familiar Integral
Z ln(2)
tan−1 (ex − 1)dx
0
[Silver]
The inverse function of this is ln(1 + tan(x)). So we have:
π
= ln(2) −
4
Z π
4
ln(1 + tan(x))
0
If you remember from a past section of sneaky queen’s rule, the following answer
π
becomes
ln(2) .
8
93
Infinite Function Example
Z 6r
q
√
3
3
x + x + 3 x + ...dx
0
[Harvard-MIT Math Tournament Integration Bee 2021 - Finals]
If we try to simplify the function algebraically, we get
q
√
3
y = x + 3 x + ...
√
y = 3x+y
y 3 − y = x.
I stopped here because solving for y is too much of a pain, but notice that I just found the
inverse of the function. It’s just a simple polynomial. So we’ll use inverse jailbreak.
If the lower bound is 0, then our new lower bound is 0......right???
Oh no....
y 3 − y = y(y − 1)(y + 1) = 0
There’s more than one solution for y, but which one is the right output???
Well, if our upper bound is 6, then our new upper bound is 2 (by inspection of the equation).
Realize that x = y 3 − y is a sleeping cubic function. Let’s test some numbers.
At y = -1 to 0, x travels forward (y = − 43 ) then back (y = − 14 ). Not okay for integration.
At y = 0 to 1, x just travels back then forward, also not okay for integration.
Then y = 1 is the output we choose because x travels in only one direction which we can
actually integrate through the interval [1,2]. Thus,
Z 2
Z 6r
q
√
3
3
3
(y 3 − y)dy
x + x + x + ...dx = 6(2) − 0 −
1
0
9
39
=
4
4
This is definitely the most dangerous inverse jailbreak integrals I’ve ever solved.
= 12 −
Logarithmic Example
√
Z 1
2
√
3
4
ln
1 − 4x2 + 1
2x
dx
[Silver]
I have no patience to do IBP on this, and I’m not risking any small mistakes. Amazingly,
ex
the inverse function of this is 1+e
2x . So we have
√
3
=−
ln(3) −
8
√
ex
π
3
dx =
−
ln(3)
√
2x
12
8
ln( 3) 1 + e
Z 0
94
Retreat from Trig-Sub
Z 1
−1
sin
q
√
1 − x dx
0
[Harvard-MIT Math Tournament Integration Bee Mock Training]
If you try x = sin4 θ or x = cos4 θ, you’re gonna have to deal with integration by parts.
Instead, the inverse function is just cos4 (x). This also gave us no leftovers!!
Z 1
sin
−1
q
1−
√
Z π
2
x dx =
cos4 (x)dx
0
0
Wallis’ Trick!!
A Very Satisfying Polynomial Inverse Example
q
Z 7 q√
3
3 √
x2 + 1 + x −
x2 + 1 − x dx
0
[Silver]
This is clearly ugly to work with. But watch very closely...
q
q
3 √
3 √
2
y=
x +1+x−
x2 + 1 − x
q
q
√
√
3 √
3 √
3
y = ( x2 + 1 + x) − 3(1)
x2 + 1 + x + 3(1)
x2 + 1 − x − ( x2 + 1 − x)
y 3 = 2x − 3y
y 3 + 3y
= x.
2
Beautiful.
If you lose track very easily,
you can do what most AIME people do,
p
p
√
√
3
3
2
let α =
x + 1 + x and let β =
x2 + 1 − x. Then simplify in terms of those variables.
So now this integral is:
Z 2 3
x + 3x
= 14 −
dx = 9
2
0
How unimaginable. This answer is just way too clean, but it turns out to be true.
Don’t believe me? Go ahead and WolframAlpha it. The amount of times I WolframAlpha
this integral just to check if this was actually true 0.0
95
6.2
Taylor Series
Taylor Series Concept
Let xg(x) be an ugly function and
g(x) =
∞
X
f (xn).
n=0
Then,
Z b
xg(x)dx =
a
∞ Z b
X
n=0
xf (nx)dx.
a
We can now integrate with an easier function and then compute the infinite sum using
taylor series or other sums.
A Beginner Example
Z 1
0
ln(1 − x)
dx
x
Turn ln(1 − x) into Taylor series:
=
Z 1X
∞
0
Z
∞
∞
X
X
−xn
1
1 1 n−1
π2
x dx = −
dx = −
=
−
nx
n 0
n2
6
n=1
n=1
n=1
Sneaky Geometric Series
Z 1
0
ln(x)
dx
1 − x2
1
Turn 1−x
2 into a Geo series:
=
Z 1X
∞
0
n=0
2n
x ln(x)dx =
∞ Z 1
X
n=0
2n
x ln(x)dx =
0
∞
X
n=0
96
−
1
π2
=
−
(2n + 1)2
8
Famous Taylor Series Log Integral
Z 1
ln(x) ln(1 − x)dx
0
Turn ln(1 − x) into a Taylor Series:
∞
X
Z
∞
∞
X
X
xn
1 1 n
1
ln(x)
− dx = −
=
x ln(x)dx =
n
n 0
n(n + 1)2
0
n=1
n=1
n=1
Z 1
=
∞
X
1
1
π2
−
=
2
−
n(n + 1) (n + 1)2
6
n=1
Exponential Case
Z ∞
ln(1 + e−x )dx
0
Z ∞X
Z
∞
∞
∞
−x n
X
X
(−1)n ∞ −x n
(−1)n
π2
n (e )
(−1)
dx =
(e ) dx =
= −
=
n
n
n2
12
0
0
n=1
n=1
n=1
Exponential Geo Series
Z ∞
0
x
dx
1 + ex
We can not use ex otherwise the bound to infinity will not converge.
So we stick with e−x .
Z ∞
Z ∞
Z ∞ X
∞
∞
X
xe−x
n
n −nx
=
dx =
x
(−1) e dx =
(−1)
xe−nx dx
−x
e +1
0
0
0
n=1
n=1
=
∞
X
(−1)n
n=1
n2
97
= −
π2
12
Taylor Series Modification
Z 1
0
x ln(x)
dx
(1 + x)2
We can also modify our Taylor Series!!
Observe that differentiating both sides gives us
∞
X
1
=
(−1)n xn
1 + x n=0
∞
X
−1
=
nxn−1 (−1)n
(1 + x)2
n=0
∞
X
−x
n(−1)n xn
=
2
(1 + x)
n=0
∞
X
x
n(−1)n xn .
=
−
(1 + x)2
n=1
After all that simplification, we can now substitute it in our integral:
Z 1
0
=
x ln(x)
dx =
(1 + x)2
Z 1
−
0
∞
X
n(−1)n
n=1
(n + 1)
=−
2
∞
X
n n
n(−1) x ln(x)dx = −
∞
X
n(−1)
n
∞
X
(n − 1)(−1)n
n2
n=2
=
∞
X
(−1)n
n=2
n2
−
xn ln(x)dx
0
n=1
n=1
Z 1
(−1)n
π2
= ln(2) −
n
12
Double Taylor Series!
Z ∞
0
Z ∞
=
ln
0
1 + e−x
1 − e−x
x
e +1
dx
ln x
e −1
Z ∞
dx =
ln(1 + e−x ) − ln(1 − e−x )dx
0
Z ∞X
Z ∞X
∞
∞
−x n
(e−x )n
n (e )
=
(−1)
dx +
dx
n
n
0
0
n=1
n=1
=
∞
X
(−1)n
n=1
n2
∞
X
1
π2
+
=
n2
4
n=1
98
Algebra Manipulation
Z 1
ln(1 + x + x2 )
dx
x
0
3
.
We can separate it into to Taylor Series since 1 + x + x2 = 1−x
1−x
By log rules, we can get
Z 1
ln(1 − x3 ) − ln(1 − x)
=
dx
x
0
=
Z 1X
∞
0
−x3n
dx +
nx
n=1
Z 1X
∞
0
xn
π2
dx =
nx
9
n=1
U-Sub Example
Z π
2
0
ln(sec(x))
dx
tan(x)
Let u = ln(sec(x)):
Z ∞
0
u
du =
tan2 (x)
Z ∞X
∞
−2u n
=
=
0
n=1
u(e
Z ∞
) du =
0
u
du =
2u
e −1
∞ Z ∞
X
n=1
0
Z ∞
0
ue−2nu du =
ue−2u
du
1 − e−2u
∞
X
1
π2
=
4n2
24
n=1
I have never seen an integral where it actually uses a “double summation” of Taylor
Series. Experimentation and creativity are the most powerful tools of problem writing!!
99
6.3
Complexifying the Integral
Beginner Concept
We have the following formulas:
sin(x) =
eix − e−ix
2i
cos(x) =
eix + e−ix
2
an − b n
= an−1 + an−2 b + an−3 b2 + ... + abn−2 + bn−1
a−b
Trust me, you’ll be surprised how many times you need this.
A Basic Example
Z 2π
0
sin(10x)
dx
sin(x)
Let’s not try to use trig-identities. Although, you could probably do periodicity, but
let’s complexify this integral anyways.
Z 2π 10ix
e
− e−10ix
=
dx
eix − e−ix
0
Let a = eix and b = e−ix . Then,
a10 − b10
= a9 + a8 b + a7 b2 + ... + ab8 + b9
a−b
Note that ab = 1, hence
= e9ix + e7ix + e5ix + ... + e−5ix + e−7ix + e−9ix
Z 2π
=
(2 cos(9x) + 2 cos(7x) + 2 cos(5x) + 2 cos(3x) + 2 cos(x))dx
0
At this point, it is really nice to use periodicity, and it is safe to say that this integral is just
0 . If you do this a lot, you’ll start to get a sense of the algebraic pattern very quickly.
100
The Significance of Periodicity
Z 2π 0
sin(3x)
sin(x)
3
dx
[Harvard-MIT Math Tournament Integration Bee 2019]
3
Z 2π
Z 2π 3ix
Z 2π
e − e−3ix
2ix
−2ix 3
(2 cos(2x) + 1)3 dx
(e + 1 + e
) dx =
=
dx =
ix − e−ix
e
0
0
0
Let u = 2x to simplify periodicity:
Z 4π
Z 4π
1 3
1
2
=
[2c(1 + cos(2x)) + 3(1 + cos(2x)) + ]du
(8c + 12c + 6c + 1)du =
2
2
0
0
Z 4π
1
3 + du = 14π
=
2
0
There was a lot of disappearing terms and cancellations thanks to periodicity.
In fact, for a very helpful tip, for every n ∈ N:
Z 2π
Z 2π
2n+1
cos2n+1 (x)dx = 0
sin
(x)dx =
0
0
No Wallis’ Trick???
Z
cos6 (x)dx
No Wallis’ Trick??? Oh naww, I’m complexifying this integral immediately!!
6
Z ix
Z
e + e−ix
1
=
dx = 6 (e6ix + 6e4ix + 15e2ix + 20 + ...)dx
2
2
Z
1
= 6 (2 cos(6x) + 12 cos(4x) + 30 cos(2x) + 20)dx
2
YAY, no double angle trig-identity algebra spamming!!
101
Advanced Concept
If you remember Euler’s complex formula, it is very well known that
eix = cos(x) + i sin(x).
Then,
Z
Z
cos(x)f (x)dx = ℜ
e f (x)dx
Z
Z
ix
sin(x)f (x)dx = ℑ
ix
e f (x)dx
A Physics Example
Z ∞
cos(x2 )dx
−∞
Z ∞
=ℜ
ix2
e
=ℜ
dx
−∞
1+i
√
2
Z ∞
e
−x2
−∞
r
dx
=
π
2
Complex Algebra Chaos!!!
Z ∞
2
e−x sin(x2 )dx
0
Z ∞
Z ∞
1
−u2
e
dx = ℑ √
=ℑ
e du
1−i 0
0
√ √
√
√
π π
1+i
π πi
π
8
√
√
√
=ℑ
=ℑ
e
=
sin
2
8
2
242
242
−x2 (1−i)
Let’s try to steer away from integrals that requires a lot of complex algebra knowledge.
No Boundaries???
Z
ecos(x) cos(x + sin(x))dx
[MIT Integration Bee 2023]
Z
=ℜ
cos(x) i(x+sin(x))
e
e
dx
Z
=ℜ
Z
ix
u
e e dx = ℜ
−ie du = ℜ −iee
eix ix
= ℜ −iecos(x)+i sin(x) = ℜ −iecos(x) (cos(sin(x)) + i sin(sin(x))) = ecos(x) sin(sin(x)) + C
102
6.4
Ninja Substitution
The following technique derives from King’s rule, except a little easier to work with
because the substitution is just u = −x.
Exponential Case
Z 1
x2
dx
x
−1 1 + e
Let u = −x:
Z 1
Z 1
x2 e x
dx
x
−1 1 + e
Z 1
Z 1 2
x (1 + ex )
1
2
dx
=
2x
dx
⇒
I
=
2I =
1 + ex
3
0
−1
x2
dx =
−x
−1 1 + e
Arctangent Case
Z 1
tan−1 (ex )dx
−1
Let u = −x:
Z 1
tan−1 (e−x )dx
−1
Z 1
2I =
−1
−1
x
−x
tan (e ) + tan (e )dx =
−1
Z 1
π
π
dx ⇒ I =
2
−1 2
A Null Case
Z π
4
ln(sec(x) − tan(x))dx
− π4
Let u = −x:
Z π
4
=
ln(sec(x) + tan(x))dx
− π4
Z π
4
2I =
ln(1)dx ⇒ I = 0
− π4
103
Logarithmic Case
Z 1
x ln(1 + ex )dx
−1
Let u = −x:
Z 1
Z 1
−x
−x ln(1 + e )dx =
=
−1
x ln
−1
Z 1
2I =
x ln(ex )dx ⇒ I =
−1
ex
ex + 1
1
3
Radical Case
Z √3
tan−1 (x)
√
dx
√
x2 + 1
− 3 x+
[UC Berkeley Integration Bee 2023]
Z √3
=
√
−1
tan
(x)(
x2 + 1 − x)dx
√
− 3
Let u = −x:
Z √3
=
√
−1
−
tan
(x)(
x2 + 1 + x)dx
√
− 3
Z √3
2I =
√
−2x tan−1 (x)dx
− 3
A Cool Gaussian Integral
Z ∞
0
tan−1 (x)
dx
xln(x)+1
[Cambridge University Integration Bee]
Let u = ln(x):
Z ∞
0
tan−1 (x)
dx =
2
xeln (x)
Let w = −u:
1
2
Z ∞
−∞
√
π −u2
π π
e du =
4
−∞ 2
Z ∞
104
2
tan−1 (eu )e−u du
dx
6.5
Inversion
A Logarithmic Case
Z ∞
0
Let u = x1 :
ln(2x)
dx
1 + x2
Z ∞
Z ∞
ln( u2 ) du
ln(2) − ln(u)
du
1 · 2 =
u2 + 1
1 + u2 u
0
0
Z ∞
ln(4)
π
du ⇒ I =
ln(2)
2I =
2
u +1
2
0
A Logarithmic Case 2
Let u = x4 :
Z ∞
ln(x)
0
x2 + 2x + 4
dx
Z ∞
Z ∞
ln( u4 ) · 4du
ln(4) − ln(u)
du
=
16
8
2
u2 + 2u + 4
( u2 + u + 4)u
0
0
Z ∞
Z ∞
ln(4)
ln(4)
π ln(2)
√
2I =
dx =
dx ⇒ I =
2
2
x + 2x + 4
(x + 1) + 3
3 3
0
0
Arctangent Case
Z ∞
0
Let u = x1 :
Z ∞
=
0
tan−1 ( u1 ) du
q 2 =
( 1 + 1) 1 u
u
Z ∞
2I =
0
tan−1 (x)
√ dx
(x + 1) x
u
Z ∞
0
tan−1 ( u1 )
√ du
(u + 1) u
tan−1 (x) + tan−1 ( x1 )
π2
√
dx ⇒ I =
4
(x + 1) x
105
Hybrid Example
Z ∞
tan−1 (x)
dx
x(ln2 (x) + 1)
0
Let u = x1 :
Z ∞
=
0
tan−1 ( x1 )
1
dx =
2
2
x(ln (x) + 1)
Z ∞
Z ∞
dx
0
(x2 + 1)(1 + ex− x )
0
π
2
π2
dx =
2
x(ln2 (x) + 1)
Exponential Example
1
Let u = x1 :
Z ∞
=
0
1
Z ∞
dx
ex− x
=
1
1 dx
(x2 + 1)(1 + e x −x )
(x2 + 1)(1 + ex− x )
0
Z
1 ∞ dx
π
=
=
2
2 0 (x + 1)
4
Needing a Proper Bound
Z ∞
1
Let u = x1 :
Z 1
=
0
ln(x)
dx
x(x − 1)
ln( u1 )
dx =
u(1 − u)
Z 1
0
ln(u)
du
u(u − 1)
Now that the bounds are from 0 to 1, we can legally perform Taylor Series!!
Double Log
Z ∞
0
Let u = x1 :
Z ∞
ln(1 − x + x2 )
dx
(1 + x2 ) ln(x)
ln(1 − x1 + x12 )
dx =
(1 + x2 ) ln( x1 )
Z ∞
ln(x2 − x + 1) − 2 ln(x)
dx
(1 + x2 ) ln( x1 )
0
0
Z ∞
Z
ln(x2 − x + 1)
1 ∞
2
π
2 ln(x)
=
−
dx =
dx =
2
2
2
(1 + x ) ln(x)
(1 + x ) ln(x)
2 0 (1 + x )
2
0
106
Dirichlet Example
Z 1
0
Let u = x1 :
sin(x) + sin( x1 )
dx
x
Z ∞
=
1
Z 1
2I =
sin(x) + sin( x1 )
x
0
sin(x) + sin( x1 )
dx
x
Z ∞
sin(x) + sin( x1 )
dx =
x
dx +
1
I=
1
2
Z ∞
0
Z ∞
0
sin(x) + sin( x1 )
dx
x
π
sin(x) sin(x)
+
dx =
x
x
2
Advanced Dirichlet Example
Z ∞
sin(x + x1 ) cos(x − x1 )
dx
x + x1
−∞
This is an even function, so we use symmetry. Then we can use u = x1 :
Z ∞
=
0
2 sin(u + u1 ) cos(u − u1 )
du =
(u + u1 )u2
Z ∞
0
sin(2u) + sin( u2 )
du
u(u2 + 1)
Separate terms and do inversion again only for the right term:
Z ∞
Z ∞
sin(2x)
x2 sin(2x)
π
sin(2x)
=
+
dx
=
dx
=
x(x2 + 1) x(x2 + 1)
x
2
0
0
Self-Substitution
Z ∞
2
1
2
xe−(x − x2 ) dx
0
[UC Berkeley Integration Bee 2023]
Let u = x1 :
Z ∞
1 −(x2 − 12 )2
x
e
dx
x3
0
√
Z ∞
Z
1 ∞ −u2
1
π
−(x2 − 12 )2
x
2I =
e
x + 3 dx =
e du ⇒ I =
x
2 −∞
4
0
=
107
A Super Cool Famous Inversion Integral
Z ∞
(x − 1)
√ x
dx
2 − 1 ln(2x − 1)
0
[JEE Main]
Now this integral is actually very tough to √
do. It’s also one of those integrals where
you can easily lose track of certain things. Let u = 2x − 1. Then,
Z ∞
Z ∞
(log2 (u2 + 1) − 1) 2udu
1
ln(u2 + 1) − ln(2)
=
·
=
du
u ln(u2 ) ln(2)
u2 + 1
(u2 + 1) ln(u)
ln2 (2) 0
0
Let u = x1 :
Z ∞
ln( x12 + 1) − ln(2)
dx =
ln2 (2)(x2 + 1) ln( x1 )
Z ∞
ln(x2 + 1) − 2 ln(x) − ln(2)
dx
− ln2 (2)(x2 + 1) ln(x)
0
0
Z ∞
Z ∞
ln(x2 + 1) − ln(2)
2
1
log2 (x2 + 1)
1
2
− 2
−
=
− 2
dx =
dx
(x2 + 1) ln(x)
x +1
ln (2)
ln2 (2)(x2 + 1) ln2 (2) (x2 + 1) ln(x)
0
0
=
Notice in my first line of the equation, the last integral form is the same as that second term
I have now, just different signs.
Z ∞
2
dx − I
I=
2
ln (2)(x2 + 1)
0
Z ∞
2
2I =
dx
2
ln (2)(x2 + 1)
0
I=
π
2 ln2 (2)
Notice how quickly and ugly it got at the very start. It is a lot to process, and it looked like
you were going nowhere at first glance. Thankfully all the ugliness cancels out nicely, giving
us a clean answer.
108
6.6
Frullani Integrals
Frullani Formula
Z ∞
0
a
f (ax) − f (bx)
dx = (f (∞) − f (0)) ln
x
b
Beginner Example
Z ∞
0
tan−1 (x) − tan−1 (πx)
dx
x
By Frullani Formula,
1
π
= lim tan (x) − lim tan (x) ln
= − ln(π)
x→0
x→∞
π
2
−1
−1
U-Sub Example
Z ∞
0
π 1
1
dx
sin
− sin
x
π
x
[CMIMC Integration Bee 2020]
Let u = x1 :
Z ∞
0
sin(u) sin(πu)
−
du =
u2
πu2
sin(x)
sin(x)
lim
− lim
x→∞
x→0
x
x
1
ln
= ln(π)
π
Algebraic Manipulation
Z ∞
0
(e3x − ex )
dx
x(ex + 1)(e3x + 1)
[MIT Integration Bee 2006]
Z ∞
=
0
1
x
1
1
− 3x
x
e +1 e +1
109
dx =
ln(3)
2
Disguised Frullani???
Z ∞
8
0
x4 + 8x
−
1
x4 + x
dx
You could do this with partial fractions or caveman, but....
Z ∞ 1
1
1
dx = ln(2)
−
x 18 x3 + 1 x3 + 1
0
Frullani Intimidation
Z ∞
0
e−xe
1 − x e−1 dx
xx+1
(x )
1
[Silver]
Z ∞ 1 −
xx
=
1
(ex)ex
x
0
A Challenging Frullani
Z ∞
0
dx =
1
1
lim x − lim x
x→∞ x
x→0 x
1
= 1
ln
e
1
1
1
sin
cos
− sec
dx
x
x
2x
[Silver]
Let u = x1 :
Z ∞
0
sin(u)(cos(u) − sec( u2 ))
du =
u2
Z ∞
=
0
1
u
0
1
u2
sin(2u) sin( u2 )
− u
2u
(2)
Another Challenging Example
Z ∞
0
Z ∞
sin(u)
sin(u) cos(u) −
du
cos( u2 )
du = − ln(4)
cos(2x) − 4 cos(x) + 3
dx
x3
[Silver]
Z ∞
(cos(2x) − 1) + 4(1 − cos(x))
dx
x3
0
Z ∞ 1
1 − cos(x)
1 − cos(2x)
=
4
−4
dx = 2 ln(2)
x
x2
(2x)2
0
=
110
6.7
Feynman Technique
Feynman Trick Concept
Here’s the easiest way to sort of explain the purpose of this trick. We have a weird
monstrous integral:
Z b
f (3g(x))
dx
g(x)
a
What we do is let a constant be a variable and derive it based from that variable. In
this case, let n = 3:
Z b
Z b
Z b
f (ng(x))
f (ng(x))
∂
′
I(n) =
f ′ (ng(x))dx
dx ⇒ I (n) =
dx =
g(x)
∂n a
g(x)
a
a
So what we did here is we simplified the integral by taking advantage of its
derivative, cancelling out the denominator. After integrating, you should get an answer
in terms of n. When you get that, integrate the n and substitute back n = 3.
Beginner Feynman Example
Z 1
0
x10 − 1
dx
ln(x)
I don’t blame you, when I was learning this technique, I was a bit uncomfortable
because you can place n wherever you want. However, the main purpose of this trick is just
to make the integral easier. One common case is utilizing:
∂ n
[x ] = xn ln(x)
∂n
Multivariables is hard to work with mentally, so experimenting is crucial. Notice with this
derivative, I can cancel out ln(x) at the bottom in the integral. Cool! I’ll do that.
Let n = 10, then:
Z 1 n
Z 1
x −1
′
I(n) =
dx ⇒ I (n) =
xn dx
ln(x)
0
0
So pretty much, we made an integral function, and then derive it in terms of n.
Now we have an easier function!!! After finding our integral function in terms of n, we integrate
it back to our original integral function.
Z 1
1
′
I (n) =
xn dx =
⇒ I(n) = ln(n + 1)
n+1
0
We had n = 10, so we just plug it back in.
Z 1 10
x −1
I(10) =
dx = ln(11)
ln(x)
0
111
A Deceiving Example
Z π
ln(2 − cos(x))dx
0
Alright, so where do we place our n??? If we try n = 2, then
1
∂
[ln(n − cos(x))] =
∂n
n − cos(x)
REMEMBER THAT FUNCTIONS of X are CONSTANTS!!!
sin(x)
in case you’re wondering. Other than that, we can definitely
So it would NOT be n−cos(x)
integrate with this derivative, so we’ll take it!!
Z π
Z π
dx
′
I(n) =
ln(n − cos(x))dx ⇒ I (n) =
0
0 n − cos(x)
Let u = tan( x2 ):
Z ∞
Z ∞
Z ∞
2du
2du
2
du
π
′
√
I (n) =
=
=
=
n−1
n + nu2 − 1 + u2
u2 (n + 1) + n − 1
n + 1 0 u2 + n+1
n2 − 1
0
0
√
−1
Then integrating this gives us I(n)
=
π
cosh
(n)
or
π
ln(n
+
n2 − 1).
√
So our answer is I(2) = π ln(2 + 3)......right???
WRONG!!! IT IS NOT!!! You must be very careful because we actually missed something.
′
The truth
√ is....when we integrate I (n), we still have to add +C. So we actually have
π ln(n + n2 − 1) + C. But how do we find C??? You must plug in another value of n
and integrate the original with it. Unfortunately, the only value we can possibly do is n = 1,
so....
Z π
Z π x dx = −π ln(2)
I(1) =
ln(1 − cos(x))dx =
ln 2 sin2
2
0
0
√
⇒ I(1) = −π ln(2) = π ln(1 + 12 − 1) + C
Ahhhhh, so C = −π ln(2), not zero. Hence, our answer is actually
I(2) = π ln(2 +
√
3) − π ln(2)
In the previous problem (the beginner example), C = 0 which most commonly
R 1 0 −1happens,
so we were able to get away with it luckily. If you let n = 0, then I(0) = 0 xln(x)
dx = 0.
Because of this, 0 = ln(0 + 1) + C → C = 0.
112
The Purpose of n Placement
Z π
sec(x) ln(1 + cos(x))dx
0
So do we let n = 1??? Well let’s see.
Z π
Z π
dx
ln(n + cos(x))
′
dx ⇒ I (n) =
I(n) =
cos(x)
0 cos(x)(n + cos(x))
0
That is not an easy integral, so we’ll retreat. As you can see, placing n isn’t about
substituting a constant, it’s about positioning it in a way where we can cancel out functions
like constants. That way, we have an easier integral.
We’ll technically still let n = 1, but if we place it differently
Z π
Z π
Z π
cos(x)
dx
ln(1 + n cos(x))
′
dx ⇒ I (n) =
dx =
I(n) =
cos(x)
0 cos(x)(1 + n cos(x))
0 1 + n cos(x)
0
Let u = tan( x2 ):
Z ∞
Z ∞
Z ∞
2du
2du
2
du
π
√
=
=
=
1+n =
2
2
2
2
1 + u + n − nu
u (1 − n) + 1 + n
1 − n 0 u + 1−n
1 − n2
0
0
π
⇒ I(n) = π sin−1 (n) + C
I ′ (n) = √
1 − n2
We can easily find C by letting n = 0 in our original integral.
Z π
ln(1 + 0)
I(0) =
dx = 0 = π sin−1 (0) + C ⇒ C = 0
cos(x)
0
Cool, we’re safe. So then our answer is
I(1) = π sin−1 (1) =
π2
2
Arctangent Example
Z ∞
0
tan−1 (2x)
dx
x(x2 + 1)
Let n = 2, then
Z ∞
tan−1 (nx)
dx ⇒ I ′ (n) =
x(x2 + 1)
Z ∞
dx
=
2
(x + 1)(n2 x2 + 1)
Z ∞
1
1
−1
n2
1
1− 12
n
−
dx
x2 + 1 x2 + n12
Z ∞
π 1 1
1
1
1 π
= 2
− 2
dx = 2
(n − 1) =
n − 1 0 x2 + n12
x +1
n −1 2
2 n+1
π
π
⇒ I(n) = ln(n + 1) + C ⇒ I(0) = 0 = ln(1) + C → C = 0
2
2
π
Hence our answer is I(2) =
ln(3) .
2
I(n) =
0
0
113
0
Arctangent Example 2
Z π
2
0
tan−1 (cos(x))
dx
cos(x)
Let n = 1 and place n in a way where we can cancel out cos(x) from the bottom.
Z π
2
I(n) =
0
tan−1 (n cos(x))
dx ⇒ I ′ (n) =
cos(x)
Z π
dx
0
n2 cos2 (x) + 1
2
Z π
Z ∞
sec2 (x)
du
1
π
√
=
dx =
=
2
2
n2 + 1 + u2
2
n2 + 1
0 n + 1 + tan (x)
0
√
π
I(n) = ln(n + n2 + 1) + C ⇒ I(0) = 0 → C = 0
2
√
π
Hence our answer is I(1) =
ln(1 + 2) .
2
2
Logarithmic Example
Z ∞
0
ln(x2 + 4)
dx
x2 + 1
Let n = 4, then
Z ∞
I(n) =
0
Z ∞
=
0
ln(x2 + n)
dx ⇒ I ′ (n) =
2
x +1
Z ∞
0
dx
(x2 + n)(x2 + 1)
√
1
1
π
n−1
π
n−1
1−n
√
√
√
+
dx
=
=
x2 + 1 x2 + n
2(n − 1)
n
2 n( n + 1)
√
I(n) = π ln( n + 1) + C ⇒ I(0) =
Z ∞
0
That integral can be easily solved by inversion.
So therefore, our answer is I(4) = π ln(3) .
114
2 ln(x)
dx = 0 → C = 0
x2 + 1
How do we make these Feynman Technique integrals???
Like always, we need to build it off from a skeleton. Let’s do some examples.
A good skeleton for this trick is a generalized integral formula. For me, I found this integral:
Z ∞
π
dx
′
√ n
=
I (n) =
n
x x −1
1
Then integrating it gives us:
Z ∞
I(n) =
1
n
2 sec−1 (x 2 )
dx = π ln(n) + C
x ln(x)
Immediately, I see a problem. How are we going to find C???
Realize that n = 0 makes the integral zero, but it also makes ln(n) undefined.
I’m very sorry to break it up with you, but this means that you can not use this integral.
Even though the Feynman trick works nicely, the process will not work.
You’re going to come across things like this A LOT, and please don’t feel bad about it. These
integrals aren’t easy to make anyways.
We know that
Z 1
1
I (n) =
x dx =
⇒ I(n) =
n+1
0
′
n
Z 1
0
xn − 1
dx
ln(x)
The reason why we have to add −1 constant on top is so that the integral can actually
converge. Without it, the whole integral will diverge. What if we use other functions like ex ?
Z 1
en − 1
′
I (n) =
exn dx =
n
0
Nevermind we can not do that o-o. That function in terms of n is not integrable.
Like I said, a lot of these integrals you’ll come across will either diverge or the process will
just not work. Even if you think you got it, please check the integral by using Desmos or
WolframAlpha to make sure it actually converges. Most of the time in my experience, it will
not.
115
6.8
Modified Famous Integrals
The easiest way to modify famous integrals is by u-substitution. However, you can
add more than one integrals together and simplify it or use integration by parts to disguise
it. You could even utilize other techniques and mix it with them. Let’s do some examples!!
An Engineering Joke
Z 1
0
(1 − x)4 x4
22
−π
dx =
2
1+x
7
Hehe...let x = tan θ, then dx = sec2 θdθ.
Z π
4
4
Z π
4
4
(1 − tan θ) tan θdθ =
0
4
Z π
4
4
(1 − tan(x)) tan (x)dx =
0
(tan(x) − tan2 (x))4 dx
0
Perfect lol.
Famous Taylor Series Integral
Z 1
ln(x) ln(1 − x)dx = 2 −
0
π2
6
We can do some algebra stuff with it.
ln2 (x) + 2 ln(x) ln(1 − x) + ln2 (1 − x) = ln2 (x − x2 )
Oh this will work nicely. We can use King’s Rule to make the integral of ln2 (x) = ln2 (1 − x),
and then Taylor’s Series.
Z 1
π2
ln2 (x − x2 )dx = 8 −
3
0
Now that is a very beautiful answer.
Fresnel Integral
Z ∞
cos(x2 )dx =
−∞
Z ∞
−∞
sin(x2 )dx =
r
π
2
Oooooh. Let’s use trig-identities!!! I like to propose this beautiful integral:
Z ∞
√
sin(x2 + x 3π + π)dx
−∞
√
You complete the square, use the angle addition trig-identity, and then you get π as your
answer!! This is literally why I enjoy making integrals so much. Now go have fun making
creative integrals!!!
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6.9
Unique Tricks
The truth is, there are infinitely many integration techniques. As ridiculous as it
sounds, it’s actually true. Do we really know 80% of the integration techniques, or did we
only discover 25%? Who know’s. But we can still keep finding them.
In this section, I will be sharing some integrals with such unique tricks that I came up
with and integrals that I found online.
A Derivative in an Integral???
Z e
1
d2 x
dx
[x ]
2
x
dx
x (1 + ln(x))
[UC Berkeley Integration Bee 2023]
d
Note that dx
[xx ] = xx (1 + ln(x)). So that means:
Z e d
[xx (1 + ln(x))]
dx =
xx (1 + ln(x))
dx
=
1
Z 2ee
1
du
= e + ln(2)
u
An Inversion Example
Z ∞
0
ln(x) ln( x1 ) + 1
dx
(x2 + 1)(ln(x) + 1)
[Stanford Math Tournament Integration Bee 2023]
Let u = x1 :
Z ∞
ln(x) ln( x1 ) + 1
dx
(x2 + 1)(ln( x1 ) + 1)
0
Z
1 ∞ ln(x) ln( x1 ) + 1
1
1
=
+
dx
2 0
(x2 + 1)
ln(x) + 1 ln( x1 ) + 1
Z
2
1 ∞ ln(x) ln( x1 ) + 1
=
dx
2 0
(x2 + 1)
ln(x) ln( x1 ) + 1
Z ∞
dx
π
=
=
2
(x + 1)
2
0
117
Yuepeng’s Ugly Trick
Z π
3
0
sec(x) + tan(x)
csc(x) + cot(x)
(sec(x) + csc(x))dx
[Silver]
Z π
3
=
sec(x)+tan(x)
eln( csc(x)+cot(x) ) (sec(x) + csc(x))dx
0
Z π
3
=
eln(sec(x)+tan(x))−ln(csc(x)+cot(x)) (sec(x) + csc(x))dx
0
Let u = ln (sec(x) + tan(x)) − ln(csc(x) + cot(x)).
Z ln( √2 +1)
3
=
−∞
2
eu du = 1 + √
3
Holy Constants
Z π
2
0
sin(2x)
p
dx
2
2
ecos (x) + e− sin (x)
[Silver]
Z π
2
=
0
1
2
1
sin(2x)e 2 sin (x)
√
dx = √
e+1
e+1
Z 1
2
0
√
2( e − 1)
2e du = √
e+1
u
Anti Conjugating
Z 1
0
√
x − x − x2
dx
2x − 1
[Harvard-MIT Math Tournament Integration Bee Mock Training]
Z 1
=
0
√
√
Z 1
x − x − x2 x + x − x2
2x2 − x
√
√
·
dx =
dx
2x − 1
x + x − x2
x − x2 )
0 (2x − 1)(x +
√
Z 1
Z 1
x
1
x
√
√
dx =
dx =
=
√
2
2
x+ 1−x
x−x
0 x+
0
The last integral can be easily solved by King’s Rule.
118
Self-Supporting U-Substitution
Z 1
√
2
ex +x + ex+ x dx
0
[AoPS Community]
Note that:
Z 1
√
x+ x
e
Z 1
dx =
0
Then
Z 1
x2 +x
e
2
eu +u 2udu
0
x2 +x
+e
Z 1
2xdx =
0
2
ex +x (2x + 1)dx
0
Z 2
=
eu du = e2 − 1
0
A Crazy Complexification
Z
x
x
ee ex cos ee
+ cos(ex ) dx
[Harvard-MIT Math Tournament Integration Bee Mock Training]
Z
ex +x
=ℜ
e
x
iee
e
iex
+e
dx
Z
=ℜ
x
e
iee +ex +x
iex +ex +x
+e
x
dx
Let u = ee and w = ex :
Z
Z
1 − i (1+i)w
iu
(i+1)w
iu
e du + e
dw = ℜ −ie +
=ℜ
e
2
1−i w
ex
= sin e + ℜ
e (cos(w) + i sin(w))
2
1 x
x
= sin(ee ) + ee (cos(ex ) + sin(ex )) + C
2
A Random ex ???
Z
x+1
√ x
dx
x xe − 1
[Silver]
Let u =
√
xex − 1:
Z
=
2udu
=
xex u
Z
√
2du
= 2 tan−1 ( xex − 1) + C
2
u +1
119
A Beautiful Sum of Inverses
Z 1
2
0
x
2
3
4
x + x + x + x + ... dx
x + x+...
[UC Berkeley Integration Bee 2022]
Notice that
y2
x
⇒
=x
y=
x+y
1−y
x=
y2
= y 2 + y 3 + y 4 + ...
1−y
Yes....both of those infinite functions are secretly inverses of each other. Cool right!!!!
1
So our answer is just
4
A Super Evil U-Substitution
Z π
4
4
3 sec (x) tan (x) + sec (x) tan3 (x) dx
0
[Silver]
Z π
4
=
4
sec4 (x) 3 tan (x) + tan3 (x) dx
0
3
Let u = 3 tan(x) + tan (x)
1
=
3
Z 4
u4 du =
0
1024
15
This Converges???
Z ∞
1
1 − x sin
dx
x
0
Z ∞
=
0
1
− sin
x
1
x
1
x
!
u= x1
Z ∞
dx −−→
0
u − sin(u)
du
u3
Integrate by parts:
Z ∞
1 sin(u) cos(u)
sin(u)
=− +
+
+
du
2
u
2u
2u
2u
0
∞
sin(u) − 2u + u cos(u)
π
π
=
+ =
2
2u
4
4
0
The rest equals to zero by Squeeze Theorem!!
120
Advanced Loophole Logging
Z ∞
0
ln(x)(ln(x) + 1)
dx
x2x−1
[Silver]
Z ∞
x
=
1−2x
Z ∞
ln(x)(ln(x) + 1)dx =
0
xe−2x ln(x) ln(x)(ln(x) + 1)dx
0
Let u = x ln(x):
Z ∞
=
ue−2u du =
0
1
4
A Sneaky Triggy U-Sub
sin2021 (x + 1)
dx
sin2023 (x)
Z
[Silver]
sin(1)
, then du = − sin
Let u = sin(x+1)
2 (x) dx.
sin(x)
−1
=
sin(1)
Z
u
2021
1
du = −
2022 sin(1)
sin(x + 1)
sin(x)
2022
+C
Holy Golden Ratio!!
Z π
2
π
3
tan(x)
p
dx
2 + tan2 (x)
[Silver]
Z π
Z ∞
Z π
2
sec(x) tan(x)
du
p
√
csc θdθ
=
dx =
=
π
u 1 + u2
sec(x) 1 + sec2 (x)
2
tan−1 (2)
3
!
π
1 + √15
1 + cos θ 2
= − ln
= ln
= ln(φ)
√2
sin θ
tan−1 (2))
5
2
121
Sneakiest Area of a Partial Circle
Z πp
2
sin(2x) sin(x)dx
0
[JEE Main]
Let u = π2 − x:
Z πp
2
sin(2x) cos(x)dx
=
0
Z πp
2
2I =
1 − (sin(x) − cos(x))2 (sin(x) + cos(x))dx
0
1
⇒I=
2
Z 1√
1 − u2 du =
−1
π
4
A Surprising Substitution!!
Z π
2
0
dx
(1 + cos(x))4
[Cambridge University]
1
Let u = 1+cos(x)
, then u12 du = sin(x)dx. Note that sin(x) =
Z 1
u4
du
·q
=
2
u
2u−1
=
1
2
Let w =
√
Z 1
√
1
2
u2
q
2
− u12 .
u
u3
du
2u − 1
2u − 1:
Z 1
=
0
12
1 2
(w + 1)3 dw =
8
35
Complex Taylor Series???
Z ∞
0
ecos(x) sin(sin(x))
dx
x
It is secretly known that
e
cos(x)
sin(sin(x)) =
∞
X
sin(nx)
n!
n=1
Hence,
=
∞ Z ∞
X
sin(nx)
n=1
0
xn!
dx =
Z
∞
X
1 ∞ sin(u)
n=1
n!
122
0
u
du =
π
(e − 1)
2
A Trippy Factorial U-Substitution
Z
x3
2
3 dx
1 + x + x2 + x6
[MIT Integration Bee 2022]
Z
x3
6
2
3 dx = 6
1 + x + x2 + x6
Z
=6
3
x2
+x+1
2
1−
2
3 dx
1 + x + x2 + x6
2
Let u = x6 + x2 + x + 1:
= 6x − 6 ln
x3 x2
+
+x+1 +C
6
2
That’s Not Linear Algebra...
Z π
2
0
(2 sin(x) + 5)
dx
(5 sin(x) + 2)2
[Nepal Integral Bee]
Z π
2
=
0
(2 sec(x) tan(x) + 5 sec2 (x))
dx =
(5 tan(x) + 2 sec(x))2
Z ∞
2
du
1
=
2
u
2
Adhesion Trick by U-Substitution???
Z √√2
3
(1 + x sin(x −
√
1 − x2 ))dx
√1
3
[Nepal Integral Bee]
Let u =
√
1 − x2 , then x =
Z √1
3
√
1 − u2 :
= √ −(1 +
2
3
√
√
udu
1 − u2 sin( 1 − u2 − u)) √
1 − u2
Z √2
3
2I = √ 2 + x sin(x −
1
√
1 − x2 ) − x sin(x −
3
r
I=
2
−
3
r
1
=
3
123
√
2−1
√
3
√
1 − x2 )dx
What About Me???
Wasabi!! (Wassup) I am Yuepeng Alex Yang, mostly known as Silver on the internet.
I’ve been speed integrating since 11th grade and started writing my own integration bee
problems in 12th grade. During the time, integration bees were not very popular, and I
wished that it was. Since there weren’t that many competitions that share their integrals,
it was hard to find newer integration bee problems. In addition, I scold any universities
who don’t share their past integration bee problems because I believe they just reuse them.
That’s where I decided to make my own integrals and recorded very cool discoveries and
unimaginable integration techniques.
I’ve actually gotten into competitive math since 7th grade and joined Mathorama.
However in my 8th grade year, I completely stopped due to bullying and the toxic elitism
in my school. I didn’t feel comfortable around any “mathletes”, so I quit pursuing in math
and began majoring in Chemistry. Because of this, I’ve never competed in an actual Math
contest in my life. I remember during high school, my math teacher would always begged me
to compete at a Math Circle, but then again, I just wasn’t comfortable being in a group of
mathletes.
During 12th grade, I debated to compete at an integration bee competition. Because
it was mostly undergrads competing, I decided to give it a try. Thankfully, it wasn’t too bad
and I was okay being there since my friends were with me. Not only that, there were rarely
any high school students competing with me; the majority were undergrads. Unfortunately
in my first integration bee competition, I got eliminated at semi-finals. I was going too
fast which made me become inaccurate with my answers; a complete lesson learned. Sadly,
integration bees are annual, so at the end of my 12th grade year, I kept coming up with more
cool integration bee problems as well as trained an amazing friend of mine who later became
top 16 qualifier for MIT Integration Bee 2022. (So proud of her!!!) During my freshmen
undergrad, I started problem writing for many integration bees and math tournaments, as
well as winning three of them in my hometown: first from Reedley College and the rest from
my own school, CSU Fresno, 2-years in a row both perfectly scored in all rounds. I wanted
to win more, but the pandemic hit sadly. Ironically, I switched late being a math major,
and thankfully finished in only two years. Now, I pursue in becoming a math professor for a
community college or a university, and currently focus on popularizing speed integration. If
no one wants to share all the dark forbidden arts of integration techniques, then I will.
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