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Coroneos' 100 Integrals
Math Ext 1 (Parramatta Marist High School)
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CORONEOS’ 100
INTEGRALS
CHALLENGE
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Preface
The art of Integration is an extremely important skill to master in
the HSC. With heavy use in Volumes, resisted motion in
Mechanics, and having its own standalone topic, if you’re not
good at Integration, you’re going to have a bad time in 4U
maths.
Coroneos’ 100 Integral challenge is known for having a rather
comprehensive coverage all the type of integrals you may
encounter in the HSC. If you are able to complete all 100
integrals with ease, then no integral in the HSC should cause
you great difficulty.
Why Does This Book Exist?
There’s no point in learning the trick/manipulation/technique/
answer to a particular question if you don’t understand how to
use it in other questions. So alongside the worked solutions,
we’ve also tried to teach you HOW to approach the question,
and questions similar to it.
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Chapter 1
The 100
Integrals
The ideal way to use this book is to:
• Do a section of ten in one go. (do NOT look at the solutions)
• Mark your answers using the quick solutions.
• Carefully read the worked solutions provided. Ask yourself:
- Was my method the most efficient one?
- Why have the solutions done what they have done?
- What would I change if I were doing this integral again?
- How should I approach this type of integral in the
future?
• Repeat the process for the next set of ten integrals.
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Section 1
Questions 1 - 10
1.
2.
x
dx
∫ x2 + 4
∫
x
dx
x2 + 4
5x + 2
3.
dx
∫ x2 − 4
4.
∫
sin x cos3 x d x
5.
∫
sin x sec3 x d x
6.
∫
cos2
7.
∫
x sin x d x
8.
∫
x sec2 2x d x
9.
∫
tan−1 2x d x
x3
dx
10.
∫ x2 + 1
x
dx
2
3
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Section 2
Questions 11 - 20
11.
x
dx
∫ (x + 2)(x + 4)
17.
12.
(x − 1)(x + 1)
dx
∫ (x − 2)(x − 3)
18.
13.
2x − 1
dx
∫ x 2 + 2x + 3
19.
x3
dx
14.
∫ 2x − 1
20.
15.
16.
∫
1+x
dx
∫ x a2 + x2
dx
∫ x a2 − x2
dx
∫ x x2 − a2
∫
x
dx
x+1
dx
1 − x − x2
dx
∫ x 2(1 − x 2) 12
4
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Section 3
Questions 21 - 30
21.
22.
∫
∫
dx
28.
∫
x+1
dx
x−1
29.
dx
∫ 5 + 3 cos x
30.
dx
∫ 3 + 5 cos x
cos−1 x
1 − x2
23.
dx
∫ x(ln x)3
24.
∫
25.
dx
∫ x 2(1 − x)
26.
dx
∫ x 2(1 + x 2)
27.
dx
∫ (1 + x 2)2
tan3 x d x
sec4 3x d x
5
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Section 4
Questions 31 - 40
sin x
dx
31.
∫ 5 + 3 cos x
32.
dx
∫ 1 + cos2 x
33.
∫ cos2 x − sin2 x
2
2
34.
∫
dx
2
x sin x d x
x2
35.
dx
∫ (x − 1)(x − 2)(x − 3)
ex
36.
dx
∫ ex − 1
37.
dx
∫ 3 sin2 x + 5 cos2 x
38.
∫
x 3e 5x −7 d x
39.
∫
x 5 ln x d x
40.
3x + 2
dx
∫ x(x + 1)3
4
6
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Section 5
Questions 41 - 50
41.
∫
ln(x 3) d x
48.
∫
42.
dx
∫ e x + e −x
49.
1 − 2x
dx
∫ 3+x
43.
∫
50.
∫
44.
dx
∫ (x 2 + 1)(x 2 + 4)
45.
∫
(x 2 + x + 1)−1 d x
46.
∫
e x sin 2x d x
47.
∫
(x 2 + x − 1)−1 d x
3
(5x 3 + 7x − 1) 2 ⋅ (15x 2 + 7) d x
1
(x 2 − x)− 2 d x
1
x 3(4 + x 2)− 2 d x
7
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Section 6
Questions 51 - 60
sin 2x
dx
51.
∫ 3 cos2 x + 4 sin2 x
sec2 x
56.
dx
∫ tan2 x − 3 tan x + 2
x2
dx
52.
∫ 1 − x4
57.
∫ (x 2 − 3x + 2) 21
53.
dx
∫ sin x cos x
58.
∫
54.
∫
59.
x
dx
∫ 1 + x3
55.
dx
∫ ex − 1
60.
∫
ln
x − 1 dx
x+1
dx
sin 2x cos x d x
x tan−1 x d x
8
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Section 7
Questions 61 - 70
61.
∫
(1 + 3x + 2x 2)−1 d x
62.
∫
(9 − x 2) 2 d x
63.
∫
(9 + x 2) 2 d x
64.
∫
x(9 + x 2) 2 d x
65.
∫
1
1
1
2
3
sec x tan x d x
66.
∫
x 2e −x d x
67.
∫
xe x d x
68.
∫
sin x tan x d x
69.
∫
sin4 x cos3 x d x
2
x3 + 1
70.
dx
∫ x3 − x
9
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Section 8
Questions 71 - 80
71.
72.
∫
ln(x +
x − 1) d x
2
77.
dx
∫ (x + 1) 12 + (x + 1)
4
73.
2
∫0
x
78.
dx
∫1 x(1 + x 2)
dx
∫0 cos2 x + 2 sin2 x
1
79.
2
74.
dx
−2 + 3x − x 2
π
2
dx
x+4
∫1
x+1
∫0
x 1 − x2 d x
4
80.
∫2
x ln x d x
2
75.
ln x
dx
∫1 x
1
76.
∫0
cos−1 x d x
10
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Section 9
Questions 81 - 90
a
(a − x)2
87.
dx
∫0 a 2 + x 2
2
dx
81.
∫1 x 2 + 5x + 4
π
2
1
−1
1
1 + sin x
82.
)
∫0 (
2
dx
∫0
x+3
dx
∫0 (x + 2)(x + 1)2
1
1
83.
88.
2 −x
x e
dx
1
7+x
84.
dx
∫0 1 + x + x 2 + x 3
x2
dx
89.
∫0 x 6 + 1
π
90.
∫0
cos2 m x d x, where m is an integer.
1
e −2x
85.
dx
∫0 e −x + 1
a
2
86.
y
dy
∫0 a − y
11
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Section 10
Questions 91 - 100
π
2
91.
∫π
4
x sin 2x d x
4
a
2
92.
∫0
π
4
93.
∫0
π
2
x2 a2 − x2 d x
∫0
98.
cos x
dx
∫0 5 − 3 sin x
1
2
sec x tan x d x
99.
(x + 2) x 2 + 4x + 5 d x
1
∫0 (4 − x 2) 23
π
2
1
94.
x2 + 4
dx
97.
∫1 x(x + 2)
100.
∫0
dx
2 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ
2
95.
∫1
x(ln x)2 d x
4
x2 + 4
96.
dx
∫3 x 2 − 1
12
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Chapter 2
Quick
Solutions
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Questions 1 - 100
WAIT!
1.
1
ln | x 2 + 4 | + C
2
2.
x2 + 4 + C
Are you sure you
3. 2 ln | x + 2 | + 3 ln | x − 2 | + C
should be here?
cos4 x
4. −
+C
4
We strongly advise you to attempt questions in blocks of ten.
5.
1
sec2 x + C
2
6.
x sin x
+
+C
2
2
And dont check your answers until you finish the block!
7. −x cos x + sin x + C
Don’t use the solutions to give you hints; you’re only cheating
yourself. Make sure you give each question your best attempt
8.
1
x
tan 2x + ln | cos 2x | + C
2
4
before coming to this section (or the next) to check your answers!
9. x tan−1 2x −
1
ln | 1 + 4x 2 | + C
4
x2 1
− ln | x 2 + 1 | + C
10.
2
2
14
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Questions 11 - 20
16. −
1 − x2
+C
x
11. 2 ln | x + 4 | − ln | x + 2 | + C
12. x + 8 ln | x − 3 | − 3 ln | x − 2 | + C
13. ln | x 2 + 2x + 3 | −
3
2
tan−1
(
2 )
x+1
1
a
+
17. − ln
a
x
a2 + x2
+C
x
1
a
18. − ln
+
a
x
a2 − x2
+C
x
+C
x3 x2 x
1
14.
+
+ +
ln | 2x − 1 | + C
6
8
8 16
1
2x + 1
+C
15. − 1 − x − x 2 + sin−1
2
)
(
5
19.
x
1
sec−1 + C
a
a
20.
2 3
1
x 2 − x + 2x 2 − 2 ln ( x + 1) + C
3
15
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Questions 21 - 30
−1
21. −
(cos x)
+C
2
x 2 − 1 + ln x +
22.
23. −
26. −
2
x2 − 1 + C
1
+C
2(ln x)2
1
− tan−1 x + C
x
x
tan−1 x
+
+C
27.
2
2(x 2 + 1)
tan2 x
28.
+ ln | cos x | + C
2
x
3
24.
tan 3x tan 3x
+
+C
3
9
25. ln
x
1
− +C
1−x
x
tan 2
1
−1
+C
29. tan
2
( 2 )
2 + tan 2x
1
30. ln
4
2 − tan 2x
+C
16
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Questions 31 - 40
1
31. − ln | 5 + 3 cos x | + C
3
36. ln | e x − 1 | + C
37.
1
32. =
2
tan−1
(
2 )
tan x
15
15
tan−1
(
3
tan x + C
5
)
+C
38.
1 5x 4−7
e
+C
20
33. ln | sec x + tan x | + C
2
34. −x cos x + 2x sin x + 2 cos x + C
35.
9
1
ln | x − 1 | − 4 ln | x − 2 | + ln | x − 3 | + C
2
2
x6
x6
ln x −
+C
39.
6
36
40. 2 ln | x | − 2 ln | x + 1 | +
2
1
−
+C
x + 1 2(x + 1)2
17
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Questions 41 - 50
ex
46.
(sin 2x − 2 cos 2x) + C
5
41. 3x ln x − 3x + C
42. tan−1(e x ) + c
47.
2
5
43. (5x 3 + 7x − 1) 2 + C
5
44.
45.
3
tan−1
(
3 )
2x + 1
ln
5
1
x
1
tan−1 x − tan−1
+C
(2)
3
6
2
1
2x + 1 −
5
2x + 1 +
5
+C
48. ln 2x − 1 + 2 x 2 − x + C
49. 7 ln | x + 3 | − 2x + C
+C
50.
1 2
(x − 8) x 2 + 4 + C
3
18
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Questions 51 - 60
56. ln | tan x − 2 | − ln | tan x − 1 | + C
51. ln(3 + sin2 x) + C
52.
1
1
1
ln | 1 + x | − ln | 1 − x | − tan−1 x + C
4
4
2
x 2 − 3x + 2 +
58. −
2
1
1
cos3 x + C OR − cos 3x − cos x + C
3
6
2
59. −
2x − 1
1
1
1
tan−1
+C
ln | 1 + x | + ln | 1 − x + x 2 | +
3
6
(
)
3
3
53. −2 ln | cosec x + cot x | + C OR = ln | tan x | + C
54.
1
1
(x − 1)ln(x − 1) − x + C
2
2
5
ln 2x − 3 + 2 x 2 − 3x + 2 + C
2
57.
55. ln | 1 − e −x | + C OR ln | e x − 1 | − x + C
1
1
x2
tan−1 x − x + tan−1 x + C
60.
2
2
2
19
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Questions 61 - 70
61. ln | 2x + 1 | − ln | x + 1 | + C
62.
63.
x
1
+ x 9 − x2 + C
9 sin−1
)
(
2[
3
]
1
9
x x 2 + 9 + ln x +
2
2
x 2 + 9 + C2
66. −e −x(x 2 + 2x + 2) + C
67.
68. ln | sec x + tan x | − sin x + C
69.
64.
1 2
3
(x + 9) 2 + C
3
1 x2
e +C
2
1 5
1
sin x − sin7 x + C
5
7
70. x − ln | x | + ln | x − 1 | + C
tan4 x
65.
+C
4
20
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Questions 71 - 80
71. x ln (x +
x2 − 1) −
x2 − 1 + C
76. 1
77.
72. 2 ln
x+1+1 +C
16
2−
73.
3 (
2)
78.
5π
2
π 2
4
1
3
1
8
74. ln
2
5
79.
1
75. (ln 2)2
2
80. 14 ln 2 − 3
21
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Questions 81 - 90
81.
82.
1
5
ln
3 (4)
2π
3 3
86.
a
(ln 4 − 1)
2
87. a(1 − ln 2)
88. ln
83. 2 − 5e −1
3
84. ln 2 + π
2
85. ln
1+e
1
− +1
( 2e ) e
89.
π
12
90.
π
2
3
+1
4
22
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Questions 91 - 100
91.
π 1
−
4 4
96. 1 +
a4
4π − 3 3 )
92.
192 (
97. 3
98.
93.
1
2
1
5
ln
3
2
99.
94.
5 5
3
3
12
(2 2 − 1)
100.
95. 2(ln 2)2 − 2 ln 2 +
5
6
ln
2
5
3
4
2
5
23
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Chapter 3
Worked
Solutions
When reading the worked solutions provided, ask yourself:
- Was my method the most efficient one?
- Why have the solutions done what they have done?
- What would I change if I were doing this integral again?
- How should I approach this type of integral in the
future?
This is the most important part when studying for Integration.
Doing the integrals themselves does absolutely nothing.
Your studying only truly begins once you carefully examine the
worked solutions and reflect upon what you would do differently
next time.
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Section 1
Questions 1 - 10
1.
x
dx
∫ x2 + 4
2.
Method 1: Reverse Chain Rule
Observing that
x
∫
dx
x2 + 4
Method 1: Reverse Chain Rule
d 2
(x + 4) = 2x:
dx
Observing that
∫
1
2x
x
d
x
=
dx
∫ x2 + 4
2 ∫ x2 + 4
x
dx =
x2 + 4
d 2
(x + 4) = 2x:
dx
1
1
2x(x 2 + 4)− 2 d x
2∫
1
1
= ln | x 2 + 4 | + C, applying the reverse chain rule.
2
1 (x 2 + 4) 2
+ C, by the reverse chain rule.
= ⋅
1
2
2
Method 2: Algebraic Substitution
Observing that
d 2
(x + 4) = 2x, u = x 2 + 4 is a suitable choice of
dx
substitution:
x
1 1
d
x
=
du
∫ x2 + 4
∫ u (2 )
=
=
=
1 du
2∫ u
2
let u = x + 4
du
= 2x
dx
1
x d x = du
2
=
x2 + 4 + C
Method 2: Algebraic Substitution #1
Observing that
d 2
(x + 4) = 2x, u = x 2 + 4 is a suitable choice of
dx
substitution:
x
1
1
d x = u− 2
du
(2 )
∫ x2 + 4
∫
=
1
ln | u | + C
2
1 −1
u 2 du
2∫
1
= u2 + C
1
ln | x 2 + 4 | + C
2
=
let u = x 2 + 4
du
= 2x
dx
1
x d x = du
2
x2 + 4 + C
25
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Method 3: Algebraic Substitution #2
= 2 sec θ + C
The goal of substitution is to eliminate difficult parts of the
integrand, thus subbing u =
∫
We then substitute back in:
x 2 + 4 will also be suitable:
x
⟹ sec θ =
tan θ =
2
u
dx =
du
∫
u
2
x +4
x
let u =
du
=
dx
= du
∫
x2 + 4
x
x2 + 4
2
x2 + 4
x
θ
2
=
2
x +4+C
x2 + 4
x d x = u du
=u+C
x2 + 4 + C
=
Method 4: Trigonometric Substitution
We can make use of the identity tan2 θ + 1 = sec2 θ to help simplify
the integrand:
∫
x
dx =
x2 + 4
=
∫
2 tan θ
(2 sec2 θ) dθ
4 tan2 θ + 4
2 tan θ
(2 sec2 θ) dθ
∫ 2 sec θ
let x = 2 tan θ
dx
= 2 sec2 θ
dθ
d x = 2 sec2 θ dθ
= 2 tan θ sec θ dθ
∫
= − 2 (cos θ)−2 (−sin θ) dθ
∫
2(cos θ)−1
=
× (−1) + C
−1
26
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3.
5x + 2
dx
∫ x2 − 4
4.
∫
sin x cos3 x d x
Method 1: Partial Fractions
Method 1: Reverse Chain Rule
Observing that the denominator has two linear factors, we shall
Observing that
use partial fraction decomposition.
5x + 2
A
B
Let 2
=
+
.
x −4
x+2 x−2
sin x cos3 x d x = − (−sin x) cos3 x d x
∫
∫
cos4 x
+ C, by the reverse chain rule.
=−
4
∴ 5x + 2 = A(x − 2) + B(x + 2)
letting x = 2: 12 = 4B ⟹ B = 3.
Method 2: Substitution
letting x = − 2: −8 = − 4A ⟹ A = 2.
Hence,
3
2
5x + 2
d
x
=
+
dx
∫(x + 2 x − 2)
∫ x2 − 4
= 2 ln | x + 2 | + 3 ln | x − 2 | + C
d
(cos x) = − sin x:
dx
Observing that
d
(cos x) = − sin x, u = cos x is a suitable
dx
substitution:
sin x cos3 x d x = u 3(−du)
∫
∫
= − u 3 du
∫
let u = cos x
du
= − sin x
dx
sin x d x = − du
u4
=−
+C
4
cos4 x
+C
=−
4
27
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5.
∫
sin x sec3 x d x
6.
∫
cos2
x
dx
2
Method 1: Reverse Chain Rule
Method 1: Half Angle Identity
Sine and cosine are very related functions, so we rewrite sec x in
terms of cos x.
Dealing with powers of trigonometric functions is not too nice, so
sin x sec3 x d x = sin x (cos x)−3 d x
∫
∫
single power:
d
Observing that
(cos x) = − sin x
dx
we will use the half-angle identity to help turn the integrand into a
1
(1 + cos 2θ):
2
As cos2 θ =
∫
= − (−sin x)(cos x)−3 d x
∫
cos2
x
1
dx =
(1 + cos x) d x
2
2∫
(cos x)−2
=−
−2
=
1
(x + sin x) + C
2
1
sec2 x + C
2
=
x sin x
+
+C
2
2
=
Method 2: Substitution
1
d
, and that
(cos x) = − sin x, hence
cos x
dx
u = cos x will be a suitable substitution:
Note that sec x =
sin x sec3 x d x = u −3 (−du)
∫
∫
u −2
+C
=−
−2
Substituting back in:
=
let u = cos x
du
= − sin x
dx
sin x d x = − du
1
sec2 x + C
2
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7.
∫
8.
x sin x d x
∫
x sec2 2x d x
Method 1: Integration by Parts
Method 1: Integration by Parts
Notice that the integrand is a product of two functions, this hints
Notice that the integrand is a product of two functions, this hints
that integration by parts will help.
that integration by parts will help.
Letting:
Letting:
u
x
−cos x
v
u
x
1
tan 2x
2
v
u′
1
sin x
v′
u′
1
sec2 2x
v′
x sin x d x = − x cos x + cos x d x
∫
∫
∫
x sec2 2x d x =
1
x
tan 2x −
tan 2x d x
2
2∫
=
x
1 sin 2x
tan 2x −
dx
2
2 ∫ cos 2x
= − x cos x + sin x + C
We then note that
d
(cos 2x) = − 2 sin 2x and we manipulate the
dx
integral such that we can apply the reverse chain rule:
=
1 (−2 sin 2x)
x
tan 2x +
dx
2
4 ∫ cos 2x
=
x
1
tan 2x + ln | cos 2x | + C
2
4
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9.
∫
tan
−1
x3
10.
dx
∫ x2 + 1
2x d x
Method 1: Integration by Parts
Method 1: Long Division
This integral is a type of integral which doesn’t succumb to normal
We observe the degree of the numerator is greater than the
integration techniques, thus we shall try integration by parts,
denominator, thus the best course of action is polynomial long
where v′ = 1.
division. This yields:
x
x3
=
x
−
x2 + 1
x2 + 1
Letting:
∫
u
tan−1 2x
x
v
u′
2
1 + 4x 2
1
v′
tan−1 2x d x = x tan−1 2x −
2x
dx
∫ 1 + 4x 2
1
2x
x3
d
x
=
x
d
x
−
dx
Hence,
∫
∫ x2 + 1
2 ∫ x2 + 1
x2 1
=
− ln | x 2 + 1 | + C
2
2
Method 2: Algebraic Manipulation
= x tan−1 2x −
8x
1
dx
4 ∫ 1 + 4x 2
Our goal with the manipulation is to obtain x 2 + 1 on the numerator
so that we are able to simplify the integral:
1
= x tan−1 2x − ln | 1 + 4x 2 | + C
4
x3
x3 + x − x
dx =
dx
∫ x2 + 1
∫ x2 + 1
x(x 2 + 1)
x
=
d
x
−
∫ x2 + 1
∫ x2 + 1
2x
1
dx
= x dx −
∫
2 ∫ x2 + 1
x2 1
− ln | x 2 + 1 | + C
=
2
2
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Section 2
Questions 11 - 20
11.
x
dx
∫ (x + 2)(x + 4)
12.
(x − 1)(x + 1)
dx
∫ (x − 2)(x − 3)
Method 1: Partial Fractions
Method 1: Partial Fractions
The degree of the numerator is the same as the denominator, so
Observing that the denominator has two linear factors, we shall
we first need to long divide. After that, our new fraction has linear
use partial fraction decomposition.
factors in the denominator, so we apply partial fractions.
Let
A
B
x
=
+
(x + 2)(x + 4)
x+2 x+4
∴ x = A(x + 4) + B(x + 2)
∴
(x − 1)(x + 1)
x2 − 1
dx =
dx
∫ (x − 2)(x − 3)
∫ x 2 − 5x + 6
letting x = − 2, 2A = − 2 ⟹ A = − 1.
x2 − 1
5x − 7
=1+ 2
Now, 2
, by long division.
x − 5x + 6
x − 5x + 6
letting x = − 4, −2B = − 4 ⟹ B = 2
Using partial fractions, let
x
−1
2
=
+
(x + 2)(x + 4)
x+2 x+4
∴ A(x − 3) + B(x − 2) = 5x − 7
letting x = 3, B = 15 − 7 ⟹ B = 8.
1
2
x
dx =
−
dx
∫(x + 4 x + 2)
∫ (x + 2)(x + 4)
= 2 ln | x + 4 | − ln | x + 2 | + C
5x − 7
A
B
=
+
.
(x − 2)(x − 3)
x−2 x−3
letting x = 2, −A = 3 ⟹ A = − 3
∴
1
1
(x − 1)(x + 1)
dx = 1dx + 8
dx − 3
dx
∫
∫ x−3
∫ x−2
∫ (x − 2)(x − 3)
= x + 8 ln | x − 3 | − 3 ln | x − 2 | + C
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2x − 1
13.
dx
∫ x 2 + 2x + 3
x3
14.
dx
∫ 2x − 1
Method 1: Algebraic Manipulation
Method 1: Long Division
Observing that
d 2
(x + 2x + 3) = 2x + 2, we shall split the
dx
Noticing that the degree of the numerator is greater than the
degree of the denominator, we shall use long division.
numerator:
1
1
1
1
x3
= x 2 + x + + 8 , by long division.
2x − 1
2
4
8 2x − 1
3
2x + 2
2x − 1
d
x
=
−
dx
∫ ( x 2 + 2x + 3 2 + (x + 1)2 )
∫ x 2 + 2x + 3
We then apply the reverse chain rule to the first fraction, and the
standard integral result to the second fraction:
2
= ln | x + 2x + 3 | −
3
2
tan
−1
(
2 )
x+1
1
1
1 2 1
x3
dx =
x + x+ +
dx
∴
∫[2
∫ 2x − 1
4
8 8(2x − 1) ]
+C
=
1
1
2
1 2 1
x + x+
dx +
dx
∫(2
4
8)
16 ∫ 2x − 1
x3 x2 x
1
=
+
+ +
ln | 2x − 1 | + C
6
8
8 16
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15.
∫
1+x
16.
dx
1 − x − x2
dx
∫ x 2(1 − x 2) 12
Method 1: Algebraic Manipulation
Method 1: Trigonometric Substitution
d
Observing that
(1 − x − x 2) = − 1 − 2x, we can split the
dx
1
1
numerator as 1 + x = − (−2x − 1) + :
2
2
We can use the Pythagorean Identity of 1 − sin2 θ = cos2 θ to help
∫
1+x
dx = −
1 − x − x2
1
2∫
−2x − 1
1 − x − x2
dx +
1
2∫
1
simplify the integral:
dx
∫ x 2(1 − x 2) 2
∫ sin2 θ 1 − sin2 θ
=
1
dθ
∫ sin2 θ
dx
1 − x − x2
1
1
1
(−1 − 2x)(1 − x − x 2)− 2 d x +
2∫
2∫
1
dx
1
5
−
x
+
(
4
2)
=−
1
=−
2
=−
1
(1 − x − x 2) 2
1
2
1
1 −1 x + 2
+ sin
5
2
let x = sin θ
dx
= cos θ
dθ
d x = cos θ dθ
= − cot θ + C
2
These integrals are now all standard integrals:
dθ
= cosec2θ dθ
∫
For our remaining integral, we then complete the square.
=−
cos θ
=
1
cos θ
+C
sin θ
We then substitute back in:
+C
x
sin θ =
⟹ cos θ =
1
1
1−x
2
x
2
θ
1 − x2
=−
1
2x + 1
+C
1 − x − x 2 + sin−1
2
)
(
5
1 − x2
+C
x
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17.
dx
∫ x a2 + x2
18.
dx
∫ x a2 − x2
Method 1: Trigonometric Substitution
Method 1: Trigonometric Substitution
We can use the Pythagorean Identity of 1 + tan2 θ = sec2 θ to help
We can use the Pythagorean Identity of 1 − sin2 θ = cos2 θ to help
simplify the integral:
simplify the integral:
dx
∫ x a2 + x2
=
a sec2 θ
∫ a tan θ a 2 + a 2 tan2 θ
let x = a tan θ
dx
= a sec2 θ
dθ
dθ
sec θ
=
dθ
∫ a tan θ
=
dx
∫ x a2 − x2
d x = a sec2 θ dθ
∫ a sin θ a 2 − a 2 sin2 θ
=−
x +a
let x = a sin θ
d x = a cos θ dθ
1
ln | cosec θ + cot θ | + C
a
x
= sin θ
a
By drawing a triangle, we can find cosec θ and cot θ:
2
dθ
By drawing a triangle, we can find cosec θ and cot θ:
1
ln | cosec θ + cot θ | + C
a
x
= tan θ
a
a cos θ
1
cosec θ dθ
=
a∫
1
cosec θ dθ
a∫
=−
=
a
x
θ
a2 − x2
2
x
θ
a
1
a
= − ln
+
a
x
1
a
= − ln
+
a
x
a2 + x2
+C
x
a2 − x2
+C
x
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19.
dx
∫ x x2 − a2
20.
∫
x
dx
x+1
Method 1: Trigonometric Substitution
Method 1: Algebraic Substitution
We can use the Pythagorean Identity of sec2 θ − 1 = tan2 θ to help
We can use the substitution u =
simplify the integral:
dx
∫ x x2 − a2
a sec θ tan θ
=
∫ a sec θ a 2 sec2 θ − a 2
dθ
=
1
dθ
a∫
let x = a sec θ
x
Now x = a sec θ ⟹ θ = sec−1 :
a
=
x
∫
let u =
x
d x = 2u du
d x = a sec θ dθ
1
= θ +C
a
u3
du
dx = 2
∫
u
+
1
x+1
x to simplify the integrand.
u3
−1
Now,
= u2 − u + 1 +
, by long division
u+1
u+1
∴
∫
x
dx = 2
x+1
∫(
u2 − u + 1 −
1
du
u + 1)
u3 u2
=2
−
+ u − ln | u + 1 | + C
2
[3
]
x
1
sec−1 + C
a
a
Now substituting back in:
=
2 3
1
x 2 − x + 2x 2 − 2 ln ( x + 1) + C
3
35
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Section 3
Questions 21 - 30
21.
∫
cos−1 x
22.
dx
1 − x2
∫
x+1
d x (no longer on standard integrals)
x−1
Method 1: Rationalising the Numerator
Method 1: Substitution
∫
We rationalise the numerator form a quadratic on the denominator:
cos−1 x
d x = − u du
∫
2
1−x
u2
+C
=−
2
−1
let u = cos x
du
1
=−
dx
1 − x2
∫
x+1
dx =
∫
x−1
=
Now substituting back in:
=
(cos−1 x)2
=−
+C
2
∫
d
(cos−1 x) = −
dx
d x = − cos−1 x −
∫
(
1 − x2
cos
−1
x
(cos−1 x)2
=−
+C
2
∫
x+1
∫
x
dx
x+1
dx
x2 − 1
dx +
x2 − 1
∫
1
dx
x2 − 1
The rightmost integral used to be a standard integral, but has
been removed from the formula sheet. Instead, we multiply top
Method 2: Reverse Chain Rule
Observe that
x+1
x+1
×
x−1
1
and bottom by x +
x 2 − 1.
. Hence:
1 − x2
1 − x2 )
1
=
dx
x2 − 1 +
(
x+
∫ x+
x2 − 1
x2 − 1
)
dx
x2 − 1
Observe that the numerator is the derivative of the denominator:
´
=
x 2 − 1 + ln x +
x2 − 1 + C
36
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23.
dx
∫ x(ln x)3
24.
Method 1: Trigonometric Identity and Reverse Chain Rule
Method 1: Reverse Chain Rule
d
1
(ln x) = :
dx
x
Observing that
We can use the Pythagorean Identity that sec2 θ = 1 + tan2 θ to
make use of the reverse chain rule:
dx
−3 1
=
(ln
x)
⋅ dx
∫ x(ln x)3 ∫
x
∫
(ln x)−2
+C
=
−2
=−
∫
sec4 3x d x
sec4 3x d x =
∫
sec2 3x(1 + tan2 3x) d x
= sec2 3x d x + sec2 3x tan2 3x d x
∫
∫
1
+C
2(ln x)2
The leftmost integral is now a standard result, so we now
Method 2: Substitution
Observing that
d
1
(ln x) = , subbing u = ln x will be suitable:
dx
x
dx
= u −3 du
∫ x(ln x)3 ∫
u −2
+C
=−
2
let u = ln x
du
1
=
dx
x
dx
= du
x
manipulate the remaining integral. Noting that
d
(tan 3x) = 3 sec2 3x, we can apply the reverse chain rule:
dx
1
(tan 3x)2(3 sec2 3x) d x
= sec2 3x d x +
∫
∫
3
tan 3x (tan 3x)3
=
+
+C
3
9
Now substituting back in:
=−
1
+C
2(ln x)2
37
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25.
dx
∫ x 2(1 − x)
26.
dx
∫ x 2(1 + x 2)
Method 1: Partial Fractions
Method 1: Trigonometric Substitution
Observing that the denominator consists of several factors, we
We can use the Pythagorean Identity 1 + tan2 θ = sec2 θ to help
can use partial fraction decomposition.
simplify the integrand.
1
A
B
C
Let 2
= + 2+
.
x (1 − x)
x x
1−x
dx
sec2 θ
=
dθ
∫ x 2(1 + x 2) ∫ tan2 θ(1 + tan2 θ)
2
= cot2 θ dθ
∫
∴ A x(1 − x) + B(1 − x) + Cx = 1
letting x = 1, C = 1
= (cosec2 θ − 1) dθ
∫
letting x = 0, B = 1
1
1
1
dx
=
+
+
dx
∫ x 2(1 − x) ∫ ( x x 2 1 − x )
= ln | x | −
1
− ln | 1 − x | + C
x
= − cot θ − θ + C
Substituting back in:
=−
= ln
d x = sec2 θ dθ
= cosec2θ dθ − dθ
∫
∫
letting x = − 1, −2A + 2B + C = 1 ⟹ A = 1
∴
let x = tan θ
dx
= sec2 θ
dθ
x
1
− +C
1−x
x
1
− tan−1 x + C
x
38
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27.
dx
∫ (1 + x 2)2
28.
∫
tan3 x d x
Method 1: Trigonometric Substitution
Method 1: Trigonometric Identity
2
2
We can use the Pythagorean Identity 1 + tan θ = sec θ to help
We can use the Pythagorean Identity 1 + tan2 θ = sec2 θ to help
simplify the integrand.
simplify the integrand.
2
dx
sec θ
=
dθ
∫ (1 + x 2)2
∫ (1 + tan2 θ)2
sec2 θ
dθ
=
∫ sec4 θ
let x = tan θ
dx
= sec2 θ
dθ
tan3 x d x = tan x(sec2 x − 1) d x
∫
∫
d x = sec2 θ dθ
= tan x sec2 x d x − tan x d x
∫
∫
= cos2 θ dθ
∫
=
1
(1 + cos 2θ) dθ
2∫
sin x
= tan x sec2 x d x −
dx
∫
∫ cos x
=
1
sin 2θ
θ+
+C
2(
2 )
= tan x sec2 x d x +
∫
∫
=
1
(θ + cos θ sin θ) + C
2
tan2 x
+ ln | cos x | + C, by the reverse chain rule.
=
2
(−sin x)
dx
cos x
By drawing a triangle, we can find cos θ and sin θ:
x
= tan θ
1
⟹ sin θ =
x
, cos θ =
x2 + 1
1
x2 + 1
x2 + 1
x
θ
1
x
tan−1 x
+
+C
=
2
2(x 2 + 1)
39
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29.
dx
∫ 5 + 3 cos x
30.
dx
∫ 3 + 5 cos x
Method 1: t-Substitution
Method 1: t-Substitution
There are no obvious/nice manipulations such substitutions, so we
There are no obvious/nice manipulations such substitutions, so we
will apply the t-substitution to convert the trigonometric terms into
will apply the t-substitution to convert the trigonometric terms into
algebraic terms:
algebraic terms:
dx
1
2
=
dt
⋅
2
3(1 − t 2 )
∫ 5 + 3 cos x ∫
1
+
t
5+
1 + t2
2
1 + t2
⋅
dt
=
∫ 5 + 5t 2 + 3 − 3t 2 1 + t 2
=2
dt
∫ 2t 2 + 8
=
dt
∫ t2 + 4
=
t
1
tan−1 + C
2
2
let t = tan
1 + t2
2
=
1 − t2
cos x =
1 + t2
2
1+t
⋅
dt
∫ 3 + 3t 2 + 5 − 5t 2 1 + t 2
=
2
dt
∫ 8 − 2t 2
=
1
dt
∫ 4 − t2
dx =
x
2
2
dt
1 + t2
1 − t2
cos x =
1 + t2
From here we need to decompose the integrand into partial
A
B
1
=
+
.
fractions. Let
4 − t2
2+t 2−t
∴ A(2 − t) + B(2 + t) = 1
Subbing back in:
=
x
let t = tan
2
2
dx =
dt
1 + t2
dx
2
1
⋅
=
dt
∫ 3 + 5 cos x ∫ 3 + 5 − 5t2 1 + t 2
tan 2x
1
tan−1
+C
2
( 2 )
letting t = 2, 4B = 1, B =
1
4
letting t = − 2, 4A = 1, A =
1
4
40
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=
1
1
1
+
dt
4 ∫(2 + t 2 − t)
=
1
(ln | 2 + t | − ln | 2 − t | ) + C
4
=
1
2+t
ln
+C
4
2−t
Now subbing back in:
2 + tan 2x
1
= ln
4
2 − tan 2x
+C
41
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Section 4
Questions 31 - 40
31.
sin x
dx
∫ 5 + 3 cos x
32.
Method 1: Reverse Chain Rule
Observing that
dx
∫ 1 + cos2 x
Method 1: t-Substitution
d
(5 + 3 cos x) = − 3 sin x:
dx
There are no nice substitutions, so we apply the t-substitution, but
before we do so, we need to remove any powers in order to
1
−3 sin x
sin x
dx = −
dx
∫ 5 + 3 cos x
3 ∫ 5 + 3 cos x
simplify the integrand:
1
= − ln | 5 + 3 cos x | + C, by the reverse chain rule.
3
Method 2: Substitution
u = cos x, u = 3 cos x, u = 5 + 3 cos x are all effective substitutions.
Applying the double result cos2 x =
1
(1 + cos 2x):
2
dx
dx
=
∫ 1 + cos2 x ∫ 1 + cos 2x + 1
2
2
However, u = 5 + 3 cos x will simplify the integrand the most:
1
1
sin x
dx =
− du
∫u( 3 )
∫ 5 + 3 cos x
=−
=−
1
ln | u | + C
3
1
ln | 5 + 3 cos x | + C
3
=
let u = 5 + 3 cos x
du
= − 3 sin x
dx
1
sin x d x = − du
3
=
2
dx
∫ cos 2x + 3
2
∫ 1 − t2 + 3
1 + t2
×
dt
1 + t2
let t = tan x
1
dx =
dt
1 + t2
1 − t2
cos 2x =
1 + t2
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=
2
∫ 1 − t2 + 3 + 3t
×
2
dt
1 + t2
33.
dx
∫ cos2 x − sin2 x
2
1 + t2
=
2
dt
∫ 2t 2 + 4
Method 1: Double Angle Identity
We observe that we can simplify the denominator using the double
1
dt
= 2
∫ t +2
=
1
=
1
angle identity for cosine: cos 2θ = cos2 θ − sin2 θ:
tan−1
(
t
tan−1
(
tan x
2
2
2)
dx
∫ cos2 x − sin2
+C
2 )
2
+C
is to multiply the top and bottom by sec x + tan x:
=
dx
sec2 x
=
dx
∫ 1 + cos2 x ∫ sec2 x + 1
Observe that
2
sec x
dx
∫ 2 + tan2 x
2
tan−1
(
d
(sec x + tan x) = sec x tan x + sec2 x, thus we apply
dx
the reverse chain rule:
d
(tan x) = sec2 x, we can apply the reverse chain rule:
dx
1
sec x(sec x + tan x)
dx
∫
sec x + tan x
sec x tan x + sec2 x
dx
=
∫
sec x + tan x
Applying the Pythagorean identity 1 + tan2 θ = sec2 θ:
=
2
dx
∫ cos x
This is now an integral we have encountered before, and the trick
We divide the top and bottom by cos2 x:
Noting that
x =
= sec x d x
∫
Method 2: Algebraic Manipulation
=
2
2 )
tan x
= ln | sec x + tan x | + C
+C
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34.
∫
x2
35.
dx
∫ (x − 1)(x − 2)(x − 3)
2
x sin x d x
Method 1: Integration by Parts
The integrand is a product of two factors, so we integrate by
Observing that the degree of the denominator is greater than the
parts.
numerator, and that denominator has three linear factors, we shall
Letting:
u
u′
∫
Method 1: Partial Fractions
x2
−cos x
2x
sin x
v
use partial fraction decomposition.
v′
x2
A
B
C
Let
=
+
+
(x − 1)(x − 2)(x − 3)
x−1 x−2 x−3
x 2 sin x d x = − x 2 cos x + 2 x cos x d x
∫
∴ A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2) = x 2
The next integral we have to deal with is another product of two
factors, so we integrate by parts again.
letting x = 2: 0 − B + 0 = 4 ⟹ B = − 4
letting x = 1: 2A + 0 + 0 = 1 ⟹ A =
1
2
letting x = 3: 0 + 0 + 2C = 9 ⟹ C =
9
2
Letting:
u
u′
∫
x
1
sin x
cos x
v
v′
1
4
9
x2
=
−
+
.
Hence,
(x − 1)(x − 2)(x − 3)
2(x − 1) x − 2 2(x − 3)
x cos x d x = x sin x − sin x d x
∫
x2
dx
Thus:
∫ (x − 1)(x − 2)(x − 3)
= x sin x + cos x + C
Hence,
∫
x 2 sin x d x = −x 2 cos x + 2(x sin x + cos x) + C
= − x 2 cos x + 2x sin x + 2 cos x + C
=
1
4
9
−
+
dx
∫ ( 2(x − 1) x − 2 2(x − 3) )
=
9
1
ln | x − 1 | − 4 ln | x − 2 | + ln | x − 3 | + C
2
2
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ex
dx
36.
∫ ex − 1
37.
Method 1: Reverse Chain Rule
Method 1: t-Substitution
dx
∫ 3 sin2 x + 5 cos2 x
There are no “nice substitutions”, so we will use the t-substitution.
d x
Observing that
(e − 1) = e x
dx
Before we do so, we need to remove all powers from the
integrand. To do so, we need to manipulate before we apply the
ex
x
|
d
x
=
ln
e
− 1 | + C, by the reverse chain rule.
∫ ex − 1
double angle identity.
dx
dx
=
∫ 3 sin2 x + 5 cos2 x ∫ 3(1 − cos2 x) + 5 cos2 x
Method 2: Substitution
Observe that
=
d x
(e − 1) = e x, so u = e x − 1 will be a suitable
dx
substitution:
ex
du
d
x
=
∫ ex − 1
∫ u
= ln | u | + C
let u = e x − 1
du
= ex
dx
e x d x = du
dx
∫ 3 + 2 cos2 x
dx
=
∫
=
dx
∫ cos 2x + 4
3 + 2(
cos 2x + 1
)
2
We are now able to carry out the t-substitution:
=
x
= ln | e − 1 | + C
1
∫ 1 − t2 + 4
×
dt
1 + t2
1 + t2
=
1
∫ 1 − t2 + 4 + 4t
×
2
dt
1 + t2
1 + t2
1 + t2
dt
=
×
∫ 3t 2 + 5 1 + t 2
let t = tan x
1
dx =
dt
1 + t2
1 − t2
cos 2x =
1 + t2
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=
1
dt
∫ 3t 2 + 5
=
1
1
dt
3 ∫ t2 + 5
38.
Method 1: Reverse Chain Rule
3
=
5)
1
3(
3
Observe that
3
t+C
5
tan−1
15
15
tan−1
(
d
(5x 4 − 7) = 20x 3, thus we manipulate such that
dx
we can apply the reverse chain rule:
∫
Substituting back in:
=
∫
4
x 3e 5x −7 d x
3
tan x + C
5
)
4
x 3e 5x −7 d x =
=
Method 2: Algebraic Manipulation
Observing that
dx
sec2 x
=
dx
∫ 3 sin2 x + 5 cos2 x ∫ 3 tan2 x + 5
d
3 tan x) =
dx (
1 5x 4−7
e
+C
20
Method 2: Algebraic Substitution
We divide the top and bottom by cos2 x:
Noting that
1
4
(20x 3)e 5x −7 d x
20 ∫
d
(5x 4 − 7) = 20x 3, u = 5x 4 − 7 will be a suitable
dx
substitution:
3 sec2 x, we manipulate such that
1
4
du
x 3e 5x −7 d x = e u
∫
∫ ( 20 )
we can apply the reverse chain rule:
=
=
3 sec2 x
1
3∫
1
15
2
=
dx
( 3 tan x) + 5
tan−1
(
1 u
e +C
20
let u = 5x 4 − 7
du
= 20x 3
dx
1
du
x3 d x =
20
Substituting back in:
3
tan x + C
5
)
=
1 5x 4−7
e
+C
20
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39.
∫
x 5 ln x d x
40.
3x + 2
dx
∫ x(x + 1)3
Method 1: Integration by Parts
Method 1: Partial Fractions
Our integrand is a product of two functions, hence we integrate by
Observing that the degree of the denominator is greater than the
parts:
numerator can contains linear factors, we decompose the
Letting:
u
u′
ln x
1
x
integrand into partial fractions.
x6
6
v
x5
v′
Let
A
B
C
D
3x + 2
=
+
+
+
x(x + 1)3
x (x + 1) (x + 1)2 (x + 1)3
∴ A(x + 1)3 + Bx(x + 1)2 + Cx(x + 1) + Dx = 3x + 2
x6
1 5
x ln x d x =
ln x −
x dx
∫
6
6∫
5
letting x = 0: A + 0 + 0 + 0 = 2 ⟹ A = 2.
letting x = − 1: 0 + 0 + 0 − D = − 1 ⟹ D = 1.
x6
1 x6
=
ln x − ⋅
+C
6
6 6
letting x = 1: 8A + 4B + 2C + D = 5
Substituting our previous values: 4B + 2C = − 12
x6
x6
ln x −
+C
=
6
36
letting x = 2: 27A + 18B + 6C + 2D = 8
Substituting our previous values: 18B + 6C = − 48
We then solve:
2B + C = − 6 and 3B + C = − 8 simultaneously.
Subtracting these equations:
(3B + C ) − (2B + C ) = − 8 − (−6) ⟹ B = − 2
Substituting this back in: C = − 2.
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Hence:
3x + 2
2
2
2
1
=
−
−
+
x(x + 1)3
x x + 1 (x + 1)2 (x + 1)3
Thus:
3x + 2
2
2
1
2
d
x
=
−
−
+
dx
∫ x(x + 1)3
∫ ( x x + 1 (x + 1)2 (x + 1)3 )
= 2 ln | x | − 2 ln | x + 1 | +
2
1
−
+C
x + 1 2(x + 1)2
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Section 5
Questions 41 - 50
41.
∫
42.
ln(x 3) d x
dx
∫ e x + e −x
Method 1: Algebraic Manipulation and Reverse Chain Rule
Method 1: Integration by Parts
We multiply the top and bottom by e x:
We can simplify the integral using the log law: loga(x b) = b loga(x):
dx
ex
=
dx
∫ e x + e −x ∫ (e x)2 + 1
ln(x 3) d x = 3 ln x d x
∫
∫
Observing
We then compute the integral using integration by parts.
Letting:
u
ln x
u′
1
x
1
= 3 x ln x − x ⋅ d x
(
∫ x )
x
1
v
v′
d x
(e ) = e x, this integral is now in a standard result:
dx
= tan−1(e x ) + C
Method 2: Algebraic Manipulation and Substitution
After the manipulation, instead of applying the reverse chain rule,
we could instead apply the substitution u = e x.
dx
ex
=
dx
∫ e x + e −x ∫ (e x)2 + 1
=
du
∫ u2 + 1
= tan
= 3 (x ln x − x) + C
= 3x ln x − 3x + C
−1
u+C
let u = e x
du
= ex
dx
e x d x = du
Substituting back in:
= tan−1(e x ) + c
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43.
∫
3
(5x 3 + 7x − 1) 2 ⋅ (15x 2 + 7) d x
44.
Method 1: Reverse Chain Rule
Observe that
Method 1: Partial Fractions (faster method)
d
(5x 3 + 7x − 1) = 15x 2 + 7, hence we can apply the
dx
reverse chain rule directly.
∫
dx
∫ (x 2 + 1)(x 2 + 4)
The denominator consists of two quadratic factors, so we
decompose the integrand into partial fractions. While the method
of substitution appears to be quite slow, we can use complex
5
3
2
(5x 3 + 7x − 1) ⋅ (15x 2 + 7) d x =
=
(5x 3 + 7x − 1) 2
5
2
numbers to speed up the process.
+C
Let
2 3
5
(5x + 7x − 1) 2 + C
5
1
A x + B Cx + D
=
+ 2
(x 2 + 1)(x 2 + 4)
x2 + 1
x +4
∴ (A x + B)(x 2 + 4) + (Cx + D)(x 2 + 1) = 1
Method 2: Algebraic Substitution
Observing that
letting x = i: (Ai + B)(−1 + 4) + 0 = 1 ⟹ Ai + B =
d
(5x 3 + 7x − 1) = 15x 2 + 7, we let
dx
Equating real and imaginary parts: B =
u = 5x 3 + 7x − 1 as our substitution.
(5x 3 + 7x − 1) 2 ⋅ (15x 2 + 7) d x = u 2 du
∫
∫
3
3
5
u2
= 5 +C
2
=
Hence:
Thus
=
1
3
1
Equating real and imaginary parts: D = − , C = 0
3
du = (15x 2 + 7) d x
Substituting back in:
1
, A=0
3
letting x = 2i: 0 + (2Ci + D)(−4 + 1) = 1 ⟹ 2Ci + D = −
let u = 5x 3 + 7x − 1
du
= 15x 2 + 7
dx
2 5
u2 + C
5
1
3
1
1
1
=
−
.
(x 2 + 1)(x 2 + 4)
3(x 2 + 1) 3(x 2 + 4)
dx
1
dx
1
dx
=
−
∫ (x 2 + 1)(x 2 + 4)
3 ∫ x2 + 1 3 ∫ x2 + 4
2 3
5
(5x + 7x − 1) 2 + C
5
=
1
x
1
tan−1 x − tan−1
+C
(2)
3
6
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Method 2: Partial Fractions
Hence:
We decompose the integrand into partial fractions:
A x + B Cx + D
1
=
+ 2
Let 2
(x + 1)(x 2 + 4)
x2 + 1
x +4
Thus
1
1
1
=
−
.
(x 2 + 1)(x 2 + 4)
3(x 2 + 1) 3(x 2 + 4)
dx
1
dx
1
dx
=
−
∫ (x 2 + 1)(x 2 + 4)
3 ∫ x2 + 1 3 ∫ x2 + 4
∴ (A x + B)(x 2 + 4) + (Cx + D)(x 2 + 1) = 1
=
1
1
x
tan−1 x − tan−1
+C
(2)
3
6
For non-4U students, we can still solve for the constants by either
equating coefficients/substituting real values. Here we choose the
method of substitution.
letting x = 0: 4B + D = 1
(1)
letting x = 1: 5A + 5B + 2C + 2D = 1
(2)
letting x = − 1: −5A + 5B − 2C + 2D = 1
(3)
letting x = 2: 16A + 8B + 10C + 5D = 1
(4)
We then solve simultaneously:
Adding equations (2) and (3): 10B + 4D = 2 (5)
Performing (5) −4 × (1): −6B = − 2 ⟹ B =
Subbing B =
1
.
3
1
4
1
back into (1): + D = 1 ⟹ D = − .
3
3
3
If we then sub these values of B and D into (2) and (4):
5A + 2C = 0 and 16A + 10C = 0
Solving these two new equations, we get that A = C = 0.
51
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45.
∫
(x 2 + x + 1)−1 d x
46.
∫
e x sin 2x d x
Method 1: Integration by Parts
Method 1: Completing the Square
Observe that we have an irreducible quadratic in the denominator,
Our integrand is a product of two functions, so we apply
integration by parts.
so we complete the square and then apply our standard integral
Letting:
result.
∫
2
−1
(x + x + 1)
dx
dx =
∫ x2 + x + 1
=
=
=
∫
∫
∫
dx
2
1
x + x + 21 + 1 − 2
2
1
(x + 2 )
2
tan−1
+
( 2 )
3
x + 21
2
3
tan−1
(
v
u′
2 cos(2x)
ex
v′
Our new integral is another product of two functions, so we apply
integration by parts again:
dx
2
u
cos(2x)
ex
v
u′
−2 sin(2x)
ex
v′
= e x sin(2x) − 2 e x cos(2x) + 2 e x sin(2x) d x
[
]
∫
+C
= e x sin(2x) − 2e x cos(2x) − 4 e x sin(2x) d x
∫
2
=
ex
e x sin 2x d x = e x sin(2x) − 2 cos(2x)e x d x
∫
3
3
sin(2x)
Letting:
1
2
u
3 )
2x + 1
Notice that we have reformed our original integral. Rearranging:
+C
5 e x sin(2x) = e x sin(2x) − 2e x cos(2x) + C1
∫
ex
Thus, e sin 2x =
(sin 2x − 2 cos 2x) + C2
∫
5
x
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47.
∫
2
(x + x − 1)
−1
1
dx
Hence:
1
=−
x2 + x − 1
Method 1: Partial Fractions
After some algebraic manipulations, the quadratic in our
Thus:
denominator is a product of two linear factors, so we decompose
∫
the integrand into partial fractions.
∫
(x 2 + x − 1)−1 d x =
∫
∫
5
+
5
x + 12 + 2
1
.
5
x + 12 − 2
dx
5∫ x+1 +
+
5
2
1
dx
5∫ x+1 −
2
5
2
dx
2
(x + 2 ) − 4
5
1
=−
5
=
5
2
1
1
(x 2 + x − 1)−1 d x = −
1
ln x +
5
5
1
1
1
+
+
+C
ln x + −
2
2
2
2
5
dx
(
5
x + 21 + 2
)(
5
x + 21 − 2
)
5
=
1
ln
x + 12 − 2
+C
5
x + 12 + 2
5
We now decompose into partial fractions.
1
=
Let 2
x +x−1
A
+
5
x + 12 + 2
B
=
1
5
5
x + 12 − 2
ln
2x + 1 −
5
2x + 1 +
5
+C
5
5
1
1
+B x+ +
=1
∴A x+ −
2
2 )
2
2 )
(
(
letting x = −
letting x = −
5
1
+
:
2
2
5B = 1 ⟹ B =
1
5
5
1
1
−
: − 5A = 1 ⟹ A = −
2
2
5
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48.
∫
1
(x 2 − x)− 2 d x (no longer on standard integrals)
49.
1 − 2x
dx
∫ 3+x
Method 1: Completing the Square
Method 1: Algebraic Manipulation
We complete the square for the quadratic in the denominator
∫
(x 2 − x)
− 21
dx =
∫
Our goal is to manipulate the numerator into a multiple of the
1
dx
denominator.
x2 − x
1 − 2x = 1 − 2(x + 3 − 3)
=
∫
1
dx
= 7 − 2(x + 3)
(x − 1)2 − 14
Hence:
This integral is now in the form of
∫
dx
7 − 2(x + 3)
1 − 2x
dx =
dx
∫
∫ 3+x
3+x
, which was
x2 − a2
=
previously a standard integral but is now removed with the
addition of the formula sheet.
= ln x −
1
+
2
7
dx − 2 dx
∫
∫ 3+x
= 7 ln | x + 3 | − 2x + C
x 2 − x + C1
= ln 2x − 1 + 2 x 2 − x + C2
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50.
∫
Method 2: Algebraic Substitution #2
1
x 3(4 + x 2)− 2 d x
Another suitable substitution is u =
x 2 + 4. Notice that this
removes the “most difficult” part of the integrand, which is the
Method 1: Algebraic Substitution #1
radical in the denominator.
Although a slightly less obvious substitution, observe that
d
(4 + x 2) = 2x, and that x 3 = x × x 2. Hence we choose
dx
∫
3
2 − 12
x (4 + x )
dx =
∫
x3
dx
4 + x2
2
u = 4 + x as our substitution.
∫
2 − 21
x 3(4 + x )
d x = (u − 4)u
∫
− 12
1
du
(2 )
1
1
1
2 − 4u − 2
u
=
) du
2 ∫(
=
1 2 3
1
u 2 − 8u 2 + C
)
2 (3
=
2
let u = x + 4
=
=
4 + x2
= (u 2 − 4) du
∫
du
= 2x
dx
1
x d x = du
2
x2 + 4
⟺ x2 = u2 − 4
du
x
=
dx
x2 + 4
du =
x
dx
x2 + 4
u3
− 4u + C
=
3
=
1 2
1
3
(x + 4) 2 − 4(x 2 + 4) 2 + C
3
⋅ x2 d x
⟺ x2 = u − 4
Substituting back in:
=
∫
x
let u =
1
u(u 2 − 12) + C
3
Substituting back in:
x2 + 4 2
(x + 4 − 12) + C
3
1 2
(x − 8) x 2 + 4 + C
3
=
1
3
=
1 2
(x − 8) x 2 + 4 + C
3
x 2 + 4(x 2 + 4 − 12) + C
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8
= − (cos θ)−3 + 8(cos θ)−1 + C
3
Method 3: Trigonometric Substitution
Another potential substitution is a trigonometric substitution,
utilising the identity 1 + tan2 θ = sec2 θ. Although not the most
=
efficient, it is still a perfectly valid method.
∫
3
2 − 21
x (4 + x )
dx =
=
∫
∫
x3
dx
let x = 2 tan θ
dx
= 2 sec2 θ
dθ
4 + x2
8 tan3 θ
2
(2 sec θ) dθ
8
8
−
+C
cos θ 3 cos3 θ
We then use a triangle to find the value of cos θ to substitute back
in:
tan θ =
d x = 2 sec2 θ dθ
4 + 4 tan2 θ
x
2
x2 + 4
⟹ cos θ =
2
x2 + 4
16 tan3 θ sec2 θ
dθ
=
∫
2 sec θ
= 8 tan3 θ sec θ dθ
∫
x
θ
2
Hence:
∫
3
2 − 12
x (4 + x )
dx = 8 ⋅
x 2 + 4 8 (x 2 + 4) x 2 + 4
− ⋅
+C
2
3
8
1
= 4 x 2 + 4 − (x 2 + 4) x 2 + 4 + C
3
sin3 θ
dθ
=8
∫ cos4 θ
=
1 2
(x − 8) x 2 + 4 + C
3
sin θ(1 − cos2 θ)
=8
dθ
∫
cos4 θ
=8
sin θ
sin θ
dθ
−
8
dθ
∫ cos2 θ
∫ cos4 θ
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Section 6
Questions 51 - 60
51.
sin 2x
dx
∫ 3 cos2 x + 4 sin2 x
Method 2: Manipulation and Substitution
Observing that
d
(sin2 x) = 2 sin x cos x = sin 2x, we can apply the
dx
Method 1: Manipulation and Reverse Chain Rule
same manipulation as in Method 1, but instead apply the
d
(sin2 x) = 2 sin x cos x = sin 2x, and that we can
Observe that
dx
substitution u = sin2 x.
2
2
change cos x = 1 − sin x.
sin 2x
2 sin x cos x
d
x
=
dx
∫ 3 cos2 x + 4 sin2 x
∫ 3(1 − sin2 x) + 4 sin2 x
2 sin x cos x
sin 2x
d
x
=
dx
∫ 3(1 − sin2 x) + 4 sin2 x
∫ 3 cos2 x + 4 sin2 x
2 sin x cos x
=
dx
∫ 3 + sin2 x
=
2 sin x cos x
dx
∫ 3 + sin2 x
=
du
∫ 3+u
= ln | 3 + u | + C
= ln | 3 + sin2 x | + C, by the reverse chain rule.
Substituting back in:
Now as 3 + sin2 x > 0 for all real x, we can remove the absolute
values:
= ln | 3 + sin2 x | + C
Now as 3 + sin2 x > 0 for all real x, we can remove the absolute
= ln(3 + sin2 x) + C
values:
= ln(3 + sin2 x) + C
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1
x2
dx
52.
∫ 1 − x4
1
1
x2
Hence:
= 4 + 4 − 22 .
2
2
(1 − x )(1 + x )
1−x 1+x x +1
Thus:
Method 1: Partial Fractions
We observe that the denominator can be factorised into linear and
quadratic factors, which we can then decompose into partial
1
dx
1
dx
1
dx
x2
d
x
=
+
−
∫ 1 − x4
4 ∫ 1 − x 4 ∫ 1 + x 2 ∫ x2 + 1
fractions.
=
1
1
1
ln | 1 + x | − ln | 1 − x | − tan−1 x + C
4
4
2
x2
x2
dx =
dx
∫ (1 − x 2)(1 + x 2)
∫ 1 − x4
x2
=
dx
∫ (1 − x)(1 + x)(1 + x 2)
A
B
Cx + D
x2
=
+
+
Let
(1 − x 2)(1 + x 2)
1−x 1+x
x2 + 1
∴ x 2 = A(1 + x)(x 2 + 1) + B(1 − x)(x 2 + 1) + (Cx + D)(1 − x 2)
letting x = i: −1 = 0 + 0 + 2Ci + 2D
1
Equating real and imaginary parts: D = − , C = 0.
2
letting x = 1: 1 = 4A ⟹ A =
1
4
letting x = − 1: 1 = 0 + 4B ⟹ B =
1
4
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53.
dx
∫ sin x cos x
Method 2: Algebraic Manipulation
If we divide the top and bottom by cos2 x:
dx
sec2 x
=
dx
∫ sin x cos x ∫ tan x
Method 1: Double Angle Formula
We use the double angle for sine: sin 2x = 2 sin x cos x to simplify
Note that
the integral:
d
(tan x) = sec2 x, so we apply the reverse chain rule:
dx
dx
2
=
dx
∫ sin x cos x ∫ 2 sin x cos x
=
= ln | tan x | + C
2 dx
∫ sin 2x
= 2 cosec x d x
∫
The integral of cosec x is another difficult one, but the
manipulation we use is to multiply the top and bottom by
cosec x + cot x.
=2
Observe that
(cosec x + cot x)cosec x
dx
∫
cosec x + cot x
d
(cosec x + cot x) = − cosec x cot x − cosec2 x, so we
dx
can apply the reverse chain rule.
= − 2 ln | cosec x + cot x | + C
59
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54.
∫
ln
55.
x − 1 dx
dx
∫ ex − 1
Method 1: Integration by Parts
We make use of the log law loga(x b) = b loga(x), and then much
like how we evaluate
∫
ln
x − 1 dx =
Letting:
∫
Method 1: Algebraic Manipulation
We divide the top and bottom of the integrand by e x:
ln x d x, we integrate by parts.
dx
e −x
=
dx
∫ e x − 1 ∫ 1 − e −x
1
ln(x − 1) d x
2∫
Observe
u
ln(x − 1)
u′
1
x−1
=
x
v
= ln | 1 − e −x | + C
1
v′
Method 2: Substitution and Partial Fractions
1
x
x ln(x − 1) −
dx
∫x−1 )
2(
The best substitution here is u = e x, in order to eliminate the “most
difficult” part of the integrand.
To evaluate the new integral, we will use an algebraic manipulation:
=
=
=
d
(1 − e −x ) = e −x, thus we apply the reverse chain rule.
dx
1
1 x−1+1
x ln(x − 1) −
dx
2
2∫ x−1
1
dx
=
du
∫ e x − 1 ∫ u(u − 1)
Our denominator now consists of two
1
1
1
x ln(x − 1) −
dx
1+
2
2 ∫(
x − 1)
linear factors, so we decompose the
let u = e x
du
= ex
dx
1
d x = du
u
integrand into partial fractions
1
1
1
x ln(x − 1) − x − ln | x − 1 | + C
2
2
2
Let
1
A
B
= +
.
u(u − 1)
u u−1
As x is restricted to x − 1 > 0 due to the domain of the integrand,
we can omit the absolute value of the logarithm:
=
1
1
(x − 1)ln(x − 1) − x + C
2
2
∴ A(u − 1) + Bu = 1
letting u = 1: B = 1 and letting u = 0: A = − 1
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Hence:
sec2 x
56.
dx
∫ tan2 x − 3 tan x + 2
1
1
1
=− +
.
u(u − 1)
u u−1
Thus:
Method 1: Substitution and Partial Fractions
du
du
dx
=
−
∫ ex − 1 ∫ u − 1 ∫ u
Note that
d
(tan x) = sec2 x, thus we choose u = tan x as our
dx
substitution.
= ln | u − 1 | − ln | u | + C
sec2 x
du
d
x
=
∫ tan2 x − 3 tan x + 2
∫ u 2 − 3u + 2
Substituting back in:
= ln | e x − 1 | − ln | e x | + C
=
= ln | e x − 1 | − x + C
du
∫ (u − 2)(u − 1)
Our denominator now consists of two linear factors, so we
decompose the integrand into partial fractions.
While this result appears different from the one obtained in Method
1, note that these two forms are equivalent when suitable log laws
are applied.
Let
1
A
B
=
+
.
(u − 2)(u − 1)
u−2 u−1
∴ A(u − 1) + B(u − 2) = 1
letting u = 1: −B = 1 ⟹ B = − 1
letting u = 2: A = 1
Hence:
1
1
1
=
−
.
(u − 2)(u − 1)
u−2 u−1
du
du
sec2 x
d
x
=
−
Thus:
∫ u−2 ∫ u−1
∫ tan2 x − 3 tan x + 2
= ln | u − 2 | − ln | u − 1 | + C
Substituting back in:
= ln | tan x − 2 | − ln | tan x − 1 | + C
61
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57.
x+1
∫ (x 2 − 3x + 2) 21
d x (no longer on standard integrals)
Method 1: Algebraic Manipulation and Reverse Chain Rule
Our goal is to manipulate the numerator into a multiple of the
d 2
(x − 3x + 2) = 2x − 3, we
derivative of the denominator. As
dx
3
1
manipulate x + 1 as so: x + 1 = (2x − 3) + + 1.
2
2
x+1
∫ (x 2 − 3x + 2) 2
dx =
1
=
1
2∫
∫
5
1
(2x
−
3)
+
2
2
58.
∫
sin 2x cos x d x
Method 1: Double Angle Identity
We apply the double angle formula for sine: sin 2x = 2 sin x cos x.
∫
sin 2x cos x d x = 2 sin x cos2 x d x
∫
We now observe that
chain rule: (alternatively you can make a substitution u = cos x)
dx
x 2 − 3x + 2
2x − 3
dx +
x 2 − 3x + 2
=−
5
2∫
d
(cos x) = − sin x, so we apply the reverse
dx
dx
2
cos3 x + C
3
Method 2: Product to Sums
x 2 − 3x + 2
Our integrand consists of a product of trigonometric functions, so
We apply the reverse chain rule to the leftmost integral, and
complete the square for the rightmost integral.
=
x 2 − 3x + 2 +
5
2∫
dx
2
3
1
(x − 2 ) − 4
=
5
3
x 2 − 3x + 2 + ln x − +
2
2
=
x 2 − 3x + 2 +
we use the product to sum identity:
1
sin A cos B = [sin(A + B) + sin(A − B)], setting A = 2x and B = x.
2
∫
sin 2x cos x d x =
x 2 − 3x + 2 + C1
1
(sin 3x + sin x) d x
2∫
=−
1
1
cos 3x − cos x + C
6
2
5
ln 2x − 3 + 2 x 2 − 3x + 2 + C2
2
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59.
x
dx
∫ 1 + x3
Thus:
x
1
dx
1
x+1
d
x
=
−
+
dx
∫ 1 + x3
3 ∫ 1 + x 3 ∫ 1 − x + x2
In order to simplify the right integral, our goal is to express the
Method 1: Partial Fractions
numerator is a multiple of the derivative of the denominator.
We can factorise the denominator into a quadratic and linear
factor, allowing us to decompose the integrand into partial
fractions.
We do so through: x + 1 =
Hence:
1
dx
1
2x − 1
1
dx
x
d
x
=
−
+
d
x
+
dx
∫ 1 + x3
3 ∫ 1 + x 6 ∫ 1 − x + x2
2 ∫ 1 − x + x2
x
x
d
x
=
dx
∫ (1 + x)(1 − x + x 2)
∫ 1 + x3
A
Bx + C
x
=
+
Now, let
(1 + x)(1 − x + x 2)
1 + x 1 − x + x2
∴ A(1 − x + x 2) + (Bx + C )(1 + x) = x
We then complete the square for the right most integral.
1
1
2x − 1
1
1
1
dx +
d
x
+
=−
dx
2
3∫ 1+x
6 ∫ 1 − x + x2
2∫
1
3
(x − 2 ) + 4
x − 21
1
1 2
1
−1
2
tan
= − ln | 1 + x | + ln | 1 − x + x | + ⋅
3
3
6
2
3
1
letting x = − 1: 3A = − 1 ⟹ A = −
3
Hence:
+C
2
1
letting x = 0: A + C = 0 ⟹ C =
3
letting x = 1: A + 2B + 2C = 1 ⟹ B =
1
3
(2x − 1) + .
2
2
1
3
=−
2x − 1
1
1
1
tan−1
+C
ln | 1 + x | + ln | 1 − x + x 2 | +
3
6
(
3
3 )
1
x+1
x
=
−
+
(1 + x)(1 − x + x 2)
3(1 + x) 3(1 − x + x 2)
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60.
∫
x tan−1 x d x
Method 1: Integration by Parts
Our integrand is a product of two functions, hence we apply
integration by parts.
Letting:
∫
x tan
−1
u
tan−1 x
1 2
x
2
v
u′
1
1 + x2
x
v′
x2
1
x2
−1
x dx =
tan x −
dx
2
2 ∫ 1 + x2
1
1
x2 + 1
x2
−1
tan x −
−
dx
=
2
2 ∫ ( 1 + x2 1 + x2 )
1
1
x2
tan−1 x −
dx
1−
=
2
2 ∫(
1 + x2 )
x2
1
1
=
tan−1 x − x + tan−1 x + C
2
2
2
64
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Section 7
Questions 61 - 70
61.
∫
62.
(1 + 3x + 2x 2)−1 d x
∫
1
(9 − x 2) 2 d x
Method 1: Partial Fractions
Method 1: Trigonometric Substitution
Our denominator can be factorised into two linear factors, so we
We make use of the Pythagorean identity 1 − sin2 θ = cos2 θ in
decompose the integrand into partial fractions.
∫
(1 + 3x + 2x 2)−1 d x =
Let
order to simplify the integral.
1
dx
∫ (2x + 1)(x + 1)
∫
(9 − x 2) 2 d x =
∫
9 − 9 sin2 θ(3 cos θ) dθ
=
∫
9 cos2 θ(3 cos θ) dθ
1
1
A
B
=
+
.
(2x + 1)(x + 1)
2x + 1 x + 1
∴ A(x + 1) + B(2x + 1) = 1
letting x = − 1: −B = 1 ⟹ B = − 1
= 9 cos2 θ dθ
∫
1 1
letting x = − : A = 1 ⟹ A = 2
2 2
=
9
(1 + cos 2θ) dθ
2∫
=
9
1
θ + sin 2θ + C
)
2(
2
=
9
9
θ + sin θ cos θ + C
2
2
Hence:
Thus:
2
1
1
=
−
(2x + 1)(x + 1)
2x + 1 x + 1
∫
(1 + 3x + 2x 2)−1 d x =
2
dx
dx −
∫ 2x + 1
∫ x+1
= ln | 2x + 1 | − ln | x + 1 | + C
let x = 3 sin θ
dx
= 3 cos θ
dθ
d x = 3 cos θ dθ
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We then substitute back in:
x
= sin θ
3
63.
3
x
θ
9 − x2
3
9−x
9
x 9 x
= sin−1 + ⋅ ⋅
2
3 2 3
9 − x2
+C
3
⟹ cos θ =
2
∫
1
(9 + x 2) 2 d x
Method 1: Trigonometric Substitution
We can make use of the Pythagorean identity 1 + tan2 θ = sec2 θ to
help simplify the integral.
∫
(9 + x 2) 2 d x =
∫
9 + 9 tan2 θ(3 sec2 θ) dθ
=
∫
9 sec2 θ(3 sec2 θ) dθ
1
1
x
=
+ x 9 − x2 + C
9 sin−1
)
(
2[
3
]
let x = 3 tan θ
dx
= 3 sec2 θ
dθ
d x = 3 sec2 θ dθ
= 9 sec3 θ dθ
∫
This integral we evaluate using integration by parts.
Letting:
∫
u
sec θ
tan θ
v
u′
sec θ tan θ
sec2 θ
v′
sec3 θ dθ = sec θ tan θ − sec θ tan2 θ dθ
∫
= sec θ tan θ − sec θ(sec2 θ − 1) dθ
∫
= sec θ tan θ − sec3 θ dθ + sec θ dθ
∫
∫
2 sec3 θ dθ = sec θ tan θ + ln | sec θ + tan θ | + C
∫
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Hence:
64.
9
9
(9 + x 2) d x = sec θ tan θ + ln | sec θ + tan θ | + C
∫
2
2
∫
1
x(9 + x 2) 2 d x
1
2
Method 1: Reverse Chain Rule
Substituting back in:
Observe that
x
= tan θ
3
x2 + 9
chain rule.
x
θ
⟹ sec θ =
x2 + 9
3
d 2
(x + 9) = 2x, hence we can apply the reverse
dx
∫
3
1
x(9 + x 2) 2 d x =
1
1
(2x)(x 2 + 9) 2 d x
2∫
3
9
= ⋅
2
x2 + 9 x 9
⋅ + ln
3
3 2
x2 + 9 x
+ C1
+
3
3
1 (x 2 + 9) 2
+C
= ⋅
3
2
2
=
1 2
3
(x + 9) 2 + C
3
1
9
x + x2 + 9
2
= x x + 9 + ln
+ C1
2
2
3
=
9
1
x x 2 + 9 + ln x +
2
2
x 2 + 9 + C2
67
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65.
∫
sec2 x tan3 x d x
66.
Method 1: Integration by Parts
Method 1: Reverse Chain Rule
Observe that
∫
x 2e −x d x
d
(tan x) = sec2 x, hence we can immediately apply
dx
Our integrand is a product of two functions, so we integrate by
parts. Letting:
the reverse chain rule.
tan4 x
2
3
+C
sec x tan x d x =
∫
4
u
x2
−e −x
v
u′
2x
e −x
v′
x 2e −x d x = − x 2e −x + 2 xe −x d x
∫
∫
Our new integral is again a product of two functions, so we
integrate by parts again. Letting:
u
x
−e −x
v
u′
1
e −x
v′
= − x 2e −x + 2 −xe −x + e −x d x
[
]
∫
= − x 2e −x − 2xe −x − 2e −x + C
= − e −x(x 2 + 2x + 2) + C
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67.
∫
2
xe x d x
68.
d 2
(x ) = 2x, so we can manipulate in order to apply
dx
the reverse chain rule.
∫
xe
x2
1
2
(2x)e x d x
dx =
2∫
=
sin x tan x d x
Method 1: Pythagorean Identity
Method 1: Reverse Chain Rule
Observe that
∫
We rewrite tan x =
sin x
and then apply the Pythagorean identity
cos x
sin2 θ + cos2 θ = 1.
sin2 x
dx
sin x tan x d x =
∫
∫ cos x
1 x2
e +C
2
1 − cos2 x
dx
=
∫ cos x
= (sec x − cos x) d x
∫
= ln | sec x + tan x | − sin x + C
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69.
∫
4
x3 + 1
70.
dx
∫ x3 − x
3
sin x cos x d x
Method 1: Pythagorean Identity and Reverse Chain Rule
We observe that
Method 1: Partial Fractions
d
(sin x) = cos x, and that the power of cos x is
dx
2
2
odd, so if we use the Pythagorean identity cos x = 1 − sin x, we
will be in a good position to apply the reverse chain rule.
We use an algebraic manipulation in order to simplify the integrand
before we decompose into partial fractions.
x3 − x + x + 1
x3 + 1
dx =
dx
∫
∫ x3 − x
x3 − x
sin4 x cos3 x d x = sin4 x(1 − sin2 x)cos x d x
∫
∫
= sin4 x cos x d x − sin6 x cos x d x
∫
∫
Now, let
1
1
= sin5 x − sin7 x + C,
5
7
by the reverse chain rule.
=
=
∫(
1+
x+1
dx
x(x 2 − 1) )
∫(
1+
1
dx
x(x − 1) )
1
A
B
= +
.
x(x − 1)
x x−1
∴ 1 = A(x − 1) + Bx
letting x = 1: B = 1
letting x = 0: A = − 1
Therefore
1
1
1
=− +
.
x(x − 1)
x x−1
1
1
x3 + 1
d
x
=
+
dx
1
−
Hence,
∫(
∫ x3 − x
x x − 1)
= x − ln | x | + ln | x − 1 | + C
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Section 8
Questions 71 - 80
71.
∫
ln (x +
x2 − 1) d x
72.
Method 1: Integration by Parts
Much like how we evaluate
Letting:
u
ln (x +
∫
∫ (x + 1) 12 + (x + 1)
Method 1: Algebraic Substitution
ln x d x, we will integrate by parts.
There are no obvious substitutions, instead we replace the “most
difficult” part of the integrand, which in this case is
x2 − 1)
x
v
1
v′
dx
∫ (x + 1) 12 + (x + 1)
1
u′
dx
x2 − 1
=
2u
du
∫ u + u2
=2
∫
ln (x +
x 2 − 1 ) d x = x ln (x +
x2 − 1) −
∫
x
du
∫ u+1
dx
let u =
x + 1.
x+1
du
1
=
dx
2 x+1
d x = 2u du
x2 − 1
= 2 ln | u + 1 | + C
We manipulate such that we can apply the reverse chain rule:
= x ln (x +
1
1
x − 1) −
(2x)(x 2 − 1)− 2 d x
2∫
2
Substituting back in:
= 2 ln
= x ln (x +
x2 − 1) −
x+1+1 +C
x2 − 1 + C
71
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4
73.
∫0
x
Method 2: Algebraic Substitution
dx
x+4
While we could substitute u = x + 4, u =
x + 4 will be a much
better substitution.
Method 1: Algebraic Manipulation
4
Our goal is to rewrite the numerator in terms of x + 4.
4
∫0
x
4
dx =
x+4
∫0
x+4−4
2 2
dx =
x+4
x+4−
=2
dx
x+4 )
4
2 3
2 3
⋅ 82 − 8 8 −
⋅ 42 − 8 4
) (3
)
(3
32 2
3
− 16 2 −
=
32 16 2
−
3
3
=
16
2−
3 (
(u 2 − 4) du
d x = 2u du
when x = 0, u = 2
2 2
4
=
∫2
16
+ 16
3
=2
16
[( 3
=2
16 8
−
(3
3
=
x+4
⟺ x = u2 − 4
du
1
=
dx
2 x+4
1 3
=2
u − 4u
[3
]
2
2
3
=
(x + 4) 2 − 8 x + 4
[3
]
0
=
let u =
u2 − 4
(2u du)
u
2 2
x+4
∫0 (
∫2
dx
4
=
∫0
x
16
2−
3 (
when x = 4, u = 2 2
2−8 2
2
)
−
8
−8
(3
)]
)
2)
2)
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2
74.
Method 2: Algebraic Substitution
dx
∫1 x(1 + x 2)
Method 1: Partial Fractions
Our denominator consists of a linear and quadratic factor, so we
use partial fractions.
We decompose into partial fractions.
Let
1
A Bx + C
=
+ 2
.
x(1 + x 2)
x
x +1
A far more obscure substitution, but one that is very succinct is
1
the substitution u = .
x
1
let
u
=
1
2
x
2
dx
1
u
=
du
−
1
du
∫1 x(1 + x 2) ∫1 1 + 1 ( u 2 )
=
−
2
u
dx
x2
1
1
u3
1
d
x
=
−
du
=
⋅ 2 du
2
u
2
∫1 u + 1 u
2
∴ 1 = A(x 2 + 1) + (Bx + C )x
1
=
letting x = 0: A = 1
u
du
∫ 1 u2 + 1
when x = 1, u = 1
1
when x = 2, u =
2
2
letting x = i: 1 = (Bi + C )i ⟹ − B + Ci = 1
Equating real and imaginary parts: B = − 1 and C = 0.
Manipulating such that we can apply the reverse chain rule:
1 1 2u
=
du
2 ∫ 1 u2 + 1
1
1
x
Therefore
=
−
x(x 2 + 1)
x x2 + 1
2
2
2
dx
x
1
Hence
=
−
dx
∫1 x(1 + x 2) ∫1 ( x x 2 + 1 )
=
1
1
2
|
|
ln
u
+
1
]1
2[
2
=
5
1
ln 2 − ln
2(
4)
=
8
1
ln
2
5
2
1
= ln | x | − ln | x 2 + 1 |
]
[
2
1
1
1
ln 5 − 0 − ln 2
) (
)
(
2
2
3
1
= ln 2 − ln 5
2
2
=
ln 2 −
=
8
1
ln
2
5
73
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1
2
∫0
ln x
75.
dx
∫1 x
76.
Method 1: Reverse Chain Rule
Method 1: Integration by Parts
Observe
1
d
(ln x) = , thus we apply the reverse chain rule.
dx
x
cos−1 x d x
This integral we deal with by integrating by parts, where v′ = 1.
Letting:
2
2
1
ln x
dx =
(ln x)2
]
[2
∫1 x
1
u′
1
= (ln 2)2 − 0
2
1
1
= (ln 2)2
2
cos−1 x
u
∫0
−1
cos
v
1
v′
1
−
1 − x2
x d x = [x cos
x
−1
1
x
0
1 − x2
1
x] +
∫
0
dx
We then manipulate such that we can apply the reverse chain rule:
= [x cos
−1
= [x cos
−1
1 1 (−2x)
x] −
dx
2 ∫0 1 − x 2
0
1
1
1
x] − [ 1 − x ]
0
2
0
= (0 − 0) − (0 − 1)
=1
74
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2
77.
∫1
π
2
x+1
78.
dx
−2 + 3x − x 2
dx
∫0 cos2 x + 2 sin2 x
Method 1: Algebraic Manipulation
Method 1: t-Substitution
Our goal is to manipulate the numerator and express it in terms of
the derivative of the bottom.
Before we apply the t-substitution, we need to simplify. We do so
1
5
As x + 1 = − (−2x + 3) + :
2
2
2
∫1
1 2
dx = −
2 ∫1
2
−2 + 3x − x
using the Pythagorean identity and double angle formula.
π
2
a
dx
dx
=
lim
dx
∫0 cos2 x + 2 sin2 x a→ π2 − ∫0 cos2 x + 2 sin2 x
2x
x+1
+
−2 + 3x − x 2
5
2∫
dx
a
−2 + 3x − x 2
= limπ −
a→ 2
We apply the reverse chain rule to the left most integral, and
complete the square to the right most integral.
2
5
2∫
= − [ −2 + 3x − x 2 ] +
1
dx
∫0 1 + 1 (1 − cos 2x)
2
a
dx
a→ 2 ∫0 3 − cos 2x
= 2 limπ −
1
dx
3
1
−
x
−
(
4
2)
2
Now we are able to apply the t-substitution.
tan a
sin−1
∫0
= 2 limπ −
2
5
= − (0 − 0) +
2
a→ 2
x − 32
1
2
tan a
∫0
= 2 limπ −
1
a→ 2
2
5
−1
= [sin (2x − 3)]
2
1
tan a
= limπ −
∫0
= limπ −
[
a→ 2
=
dx
5
−1
−1
(sin 1 − sin (−1))
2
a→ 2
5π
=
2
1
1 + t2
2
3 − 1 − t2
1+t
let t = tan x
1
dx =
dt
1 + t2
dt
1
dt
2 + 4t 2
when x = 0, t = 0
when x = a, t = tan a
1
dt
1 + 2t 2
tan a
tan−1 ( 2t)
]
2
1
0
75
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=
1
1
limπ − tan−1 ( 2 tan a)
79.
2 a→ 2
Now as a →
π−
,
2
2 tan a → + ∞, hence tan−1 ( 2 tan a) →
=
1
⋅
2
=
π
2
π−
.
2
∫0
x 1 − x2 d x
Method 1: Reverse Chain Rule
Observe that
d
(1 − x 2) = − 2x, so we manipulate such that we
dx
can apply the reverse chain rule.
π 2
1
1 1
x 1 − x dx = −
(−2x) 1 − x 2 d x
∫0
2 ∫0
4
2
Method 2: Algebraic Manipulation
π
2
a
dx
dx
=
lim
dx
∫0 cos2 x + 2 sin2 x a→ π2 − ∫0 cos2 x + 2 sin2 x
=−
We divide the top and bottom by cos2 x.
1
= − (0 − 1)
3
a
sec2 x
dx
= limπ −
a→ 2 ∫0 1 + 2 tan2 x
= limπ −
a→ 2
= limπ −
a→ 2
=
1
1
a
2 ∫0
=
2 sec2 x
1
1 2
3
⋅ [(1 − x 2) 2 ]
2 3
0
1
3
dx
2
1 + ( 2 tan x)
a
tan−1 ( 2 tan x)
]
[
2
1
0
limπ − tan−1 ( 2 tan a)
2 a→ 2
=
π 2
4
(by the same argument above)
76
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4
80.
∫2
x ln x d x
Method 1: Integration by Parts
Our integrand is a product of two integrals, so we apply integration
by parts. Letting:
u
ln x
1 2
x
2
v
u′
1
x
x
v′
4
1 4
1 2
x ln x −
x dx
x ln x d x =
]
[2
∫2
∫2
2
2
4
4
1 x2
16
ln 4 − ln 2 −
=
) 2[2]
( 2
2
4
2
1
= (16 ln 2 − 2 ln 2) − (8 − 2)
2
= 14 ln 2 − 3
77
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Section 9
Questions 81 - 90
2
81.
Hence:
dx
∫1 x 2 + 5x + 4
2
2
dx
1
1
=
+
dx
−
∫1 x 2 + 5x + 4 ∫1 ( 3(x + 4) 3(x + 1) )
Method 1: Partial Fractions
Factorising the denominator into two linear factors, we then
decompose the integrand into partial fractions.
2
dx
dx
=
∫1 x 2 + 5x + 4 ∫1 (x + 4)(x + 1)
=
1
2
[ln | x + 1 | − ln | x + 4 | ]1
3
=
1
[(ln 3 − ln 2) − (ln 6 − ln 5)]
3
=
1
3×5
ln
3 (2 × 6)
=
1
5
ln
3 (4)
2
Let
1
A
B
=
+
.
(x + 4)(x + 1)
x+4 x+1
∴ A(x + 1) + B(x + 4) = 1
letting x = − 1: 3B = 1 ⟹ B =
1
3
letting x = − 4: −3A = 1 ⟹ A = −
Therefore
1
3
1
1
1
=−
+
.
(x + 4)(x + 1)
3(x + 4) 3(x + 1)
78
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π
2
−1
1
82.
1 + sin x
)
∫0 (
2
dx
=
4
tan−1
3
Method 1: t-Substitution
=
There are no immediate substitutions or manipulations, and our
integrand contains a single trigonometric function, hence we apply
=
the t-substitution.
π
2
π
2
1
2
(1 + sin x)−1 d x =
dx
∫0
∫0 2 + sin x
2
=
tan−1
3(
4
(
3 )
1
2t + 1
3 − tan−1
0
3)
1
π π
−
(
3
6)
3
4
2π
3 3
1
=
∫0 2 +
2
2t
1 + t2
2
dt
1 + t2
×
1
=2
dt
∫0 1 + t + t 2
We then complete the square for the quadratic in the denominator:
1
=
∫0 3
4
2
2
dt
+ (t + 21 )
1
=
4
3
tan−1
t + 12
3
2
0
79
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1
1
83.
∫0
2 −x
x e
84.
dx
Method 1: Integration by Parts
Method 1: Partial Fractions
Our integrand is a product of two functions, so we apply
integration by parts. Letting:
1
∫0
2 −x
x e
Factorising the denominator into a linear and quadratic factor, we
then decompose the integrand into partial fractions.
u
x2
−e −x
v
u′
2x
e −x
v′
1
d x = [−x 2e −x] + 2
0
∫
7+x
dx
∫0 1 + x + x 2 + x 3
1
1
7+x
7+x
d
x
=
dx
∫0 1 + x + x 2 + x 3
∫0 (x + 1)(x 2 + 1)
Now, let
1
xe −x d x
A
Bx + C
7+x
=
+
.
(x + 1)(x 2 + 1)
x+1
x2 + 1
∴ A(x 2 + 1) + (Bx + C )(x + 1) = 7 + x
0
Our new integrand is again a product of two functions, so we
integrate by parts again. Letting:
letting x = − 1: 2A = 6 ⟹ A = 3
letting x = 0: A + C = 7 ⟹ C = 4
u
x
−e −x
v
u′
1
e −x
v′
letting x = 1: 2A + 2B + 2C = 8 ⟹ B = − 3
Therefore
7+x
3
−3x + 4
=
+
(x + 1)(x 2 + 1)
x+1
x2 + 1
1
= [− x e ] + 2 [ − xe ] + e −x d x
∫
]
[
0
0
2 −x
1
1
−x
1
1
7+x
3x
4
3
Hence:
d
x
=
−
+
dx
∫0 1 + x + x 2 + x 3
∫0 ( x + 1 x 2 + 1 x 2 + 1 )
0
1
3
= 3 ln | x + 1 | − ln | x 2 + 1 | + 4 tan−1 x
]
[
2
0
1
= [− x 2e −x − 2xe −x − 2e −x]
0
= − [e (x + 2x + 2)]
−x
2
1
=
0
3
ln 2 + π
2
= 2 − 5e −1
80
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1
e −2x
85.
dx
∫0 e −x + 1
1
1
1
1
=
−
+
+
u 2(1 + u)
u u2 u + 1
Therefore:
1
e
e −2x
1
1
1
Hence:
d
x
=
+
+
du
−
∫0 e −x + 1
∫1 ( u u 2 u + 1 )
Method 1: Substitution and Partial Fractions
There are no immediate substitutions/manipulations, so a good
substitution to try would be u = e x.
1
e
e −2x
u −2
dx =
⋅ u −1 du
−x
−1
∫0 e + 1
∫1 u + 1
e
=
u
−3
∫1 u −1 + 1
du
e
=
du
∫1 u 3 + u 2
e
let u = e x
dx
= ex
du
1
d x = du
u
1+u
1
= ln
−
u
u]
[
1
=
ln
(
when x = 0, u = 1
when x = 1, u = e
e
=
e
1
= −ln u − + ln | 1 + u |
]
[
u
1
du
∫1 u 2(1 + u)
= ln
1+e
1
−
− (ln 2 − 1)
e
e)
1+e
1
− +1
( 2e ) e
Our denominator consists of a linear and quadratic factor, so we
decompose into partial fractions.
Let
1
A
B
C
=
+
+
.
u 2(1 + u)
u u2 u + 1
∴ 1 = Au(u + 1) + B(u + 1) + Cu 2
letting u = 0: B = 1
letting u = − 1: C = 1
letting u = 1: 1 = 2A + 2B + C ⟹ A = − 1
81
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a
2
a
y
86.
dy
∫0 a − y
(a − x)2
87.
dx
∫0 a 2 + x 2
Method 1: Algebraic Manipulation
Method 1: Algebraic Manipulation
Our goal is to express the numerator in terms of the denominator.
Expanding the numerator, we can manipulate the integral such
As y = − (a − y) + a:
that we can easily apply the reverse chain rule.
a
2
a
2
y
a−y
a
dy = −
dy +
dy
∫0 a − y
∫0 a − y
∫0 a − y
a
2
=−
∫0
1 dy − [a ln | a − y | ]
a
2
a
=
0
a
2
a
2
0
0
= − [y] − [a ln | a − y | ]
=−
a 2
a − 2a x + x 2
(a − x)2
dx =
dx
∫0
∫0 a 2 + x 2
a2 + x2
a
a
2
2a x
dx
a2 + x2 )
= [x − a ln | a 2 + x 2 | ]
a
a
− a ln
− ln | a |
2
2
(
)
a
0
= (a − a ln | 2a 2 | ) − (0 − a ln | a 2 | )
(2)
a
= − − a ln
2
a
a
=−
∫0 (
1−
2a 2
= a − a ln
a2
a
1
− a ln
(2)
2
= a(1 − ln 2)
a
= − + a ln 2
2
a
= (ln 4 − 1)
2
82
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1
1
x+3
88.
dx
∫0 (x + 2)(x + 1)2
x2
dx
89.
∫0 x 6 + 1
Method 1: Partial Fractions
Method 1: Reverse Chain Rule
Our denominator consists of a linear and quadratic factor, so we
Observe that
decompose the integrand into partial fractions.
d 3
(x ) = 3x 2, thus we can manipulate such that we
dx
can apply the reverse chain rule.
A
B
C
x+3
=
+
+
.
Let
(x + 2)(x + 1)2
x + 2 x + 1 (x + 1)2
1
x2
1 1 3x 2
dx =
dx
∫0 x 6 + 1
3 ∫0 (x 3)2 + 1
∴ A(x + 1)2 + B(x + 1)(x + 2) + C(x + 2) = x + 3
letting x = − 1: C = 2
=
1
1
−1 3
tan
(x
)
]0
3[
=
1 π
−0
)
3 (4
=
π
12
letting x = − 2: A = 1
letting x = 0: A + 2B + 2C = 3 ⟹ B = − 1
x+3
1
1
2
=
−
+
(x + 2)(x + 1)2
x + 2 x + 1 (x + 1)2
Therefore
1
1
2
1
x+3
d
x
=
−
+
dx
Hence:
∫0 ( x + 2 x + 1 (x + 1)2 )
∫0 (x + 2)(x + 1)2
1
1
2
= ln | x + 2 | − ln | x + 1 | −
[
x + 1 ]0
= (ln 3 − ln 2 − 1) − (ln 2 − 0 − 2)
= ln
3
+1
4
83
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π
90.
∫0
cos2 m x d x, where m is an integer.
Method 1: Half Angle Formula
We aim to simplify the integral by reducing the power of cos(m x).
1
We do so by applying the half angle formula cos2 x = (1 + cos 2x).
2
π
1 π
(1 + cos 2m x) d x
cos m x d x =
∫0
2 ∫0
2
π
sin 2m x
1
x−
=
2[
2m ]0
=
1
1
π−
sin(2mπ) − 0
)
2(
2m
Now the sine of any integer multiple of π is zero, hence:
π
∫0
cos2 m x d x =
π
2
84
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Section 10
Questions 91 - 100
a
2
π
2
91.
∫π
92.
x sin 2x d x
4
Method 1: Integration by Parts
∫0
x2 a2 − x2 d x
Method 1: Trigonometric Substitution
Our integrand is a product of two functions, hence we integrate by
We can use the Pythagorean identity 1 − sin2 θ = cos2 θ to help
parts. Letting:
simplify the integral.
u
x
u′
1
−
π
2
π
4
4
1
cos 2x
2
v
sin 2x
v′
let x = a sin θ
dx
= a cos θ
dθ
d x = a cos θ dθ
π
2
1 2
1
x sin 2x d x = − x cos 2x +
cos 2x d x
]π 2 ∫π
[ 2
∫π
4
π
2
1
1
= − x cos 2x + sin 2x
]π
[ 2
4
a
2
∫0
x2 a2 − x2 d x =
4
=
=
π
π
+0 − 0−
) (
(4
4)
π 1
−
4 4
when x = 0, θ = 0
π
when x = a, θ =
6
π
6
∫0
= a4
(a sin θ)2 a 2 − a 2 sin2 θ(a cos θ) dθ
π
6
∫0
sin2 θ cos2 θ dθ
We then use the double and half angle formula to help simplify the
integral.
85
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π
4
π
a4 6
=
(2 sin θ cos θ)2 dθ
4 ∫0
93.
∫0
sec2 x tan x d x
π
a4 6 2
=
sin 2θ dθ
4 ∫0
Method 1: Reverse Chain Rule
π
a4 6
(1 − cos 4θ) dθ
=
8 ∫0
π
6
a4
1
=
θ − sin 4θ
]
8 [
4
0
Observe that
d
(tan x) = sec2 x, hence we can immediately apply
dx
the reverse chain rule.
π
π
4
4
1
2
2
tan x
sec x tan x d x =
]
[2
∫0
0
a4 π 1
2π
=
− sin
−0
8 (6 4
3 )
3
a4 π
=
−
8 (6
8 )
=
1
−0
(2
)
=
1
2
a4
4π − 3 3 )
=
192 (
86
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2
1
94.
∫0
2
95.
(x + 2) x + 4x + 5 d x
Method 1: Integration by Parts
Method 1: Reverse Chain Rule
Observe that
∫1
x(ln x)2 d x
d 2
(x + 4x) = 2x + 4, hence we manipulate such
dx
Our integrand is a product of two functions, hence we integrate by
parts. Letting:
that we can apply the reverse chain rule.
1
1
1
(2x + 4) x 2 + 4x + 5 d x
(x + 2) x + 4x + 5 d x =
∫0
2 ∫0
u
(ln x)2
x2
2
v
u′
2 ln x
x
x
v′
2
1
1 2
3
2
2
= ⋅ [(x + 4x + 5) ]
2 3
0
=
1
3
3
2 − 52
10
)
3(
1
= (10 10 − 5 5 )
3
=
5 5
3
2
2
x2
2
2
(ln x)
− x ln x d x
x(ln x) d x =
∫1
] ∫1
[2
2
1
Our new integrand is again a product of two functions, hence we
integrate by parts again. Letting:
(2 2 − 1)
u
ln x
x2
2
v
u′
1
x
x
v′
2
2
1 2
1 2
x2
2
(ln x)
−
x ln x −
x dx
=
]
∫1
2
] ([ 2
)
[2
1
1
87
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4
2
2
x2 + 4
dx
96.
∫3 x 2 − 1
1
1
1 2
x (ln x)2 − x 2 ln x − x 2
x(ln x) d x =
[2
∫1
2
4 ]1
2
1
1
1
1
=
⋅ 4(ln 2)2 − ⋅ 4 ln 2 − ⋅ 4 − 0 − 0 −
[2
2
4 ] (
4)
3
= 2(ln 2) − 2 ln 2 +
4
2
Method 1: Partial Fractions
We simplify by rewriting the numerator in terms of the
denominator, and then decomposing into partial fractions.
4
x2 + 4
5
x2 − 1
d
x
=
+
dx
∫3 x 2 − 1
∫3 ( x 2 − 1 x 2 − 1 )
4
Now, let
A
B
1
=
+
.
x2 − 1
x+1 x−1
∴ A(x − 1) + B(x + 1) = 1
letting x = 1: 2B = 1 ⟹ B =
1
2
letting x = − 1: −2A = 1 ⟹ A = −
1
2
1
1
1
=
−
+
x2 − 1
2(x + 1) 2(x − 1)
Therefore
4
4
x2 + 4
1
1
Hence:
d
x
=
+
dx
1
−
∫3 x 2 − 1
∫3 (
2(x + 1) 2(x − 1) )
4
1
1
= x − ln | x + 1 | + ln | x − 1 |
[
]
2
2
3
=
(
4−
=1+
1
1
1
1
ln 5 + ln 3 − 3 − ln 4 + ln 2
) (
)
2
2
2
2
6
5
ln
2
5
88
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4
x2 + 4
97.
dx
∫1 x(x + 2)
Hence:
4
4
x2 + 4
4
2
dx =
dx
1+ −
∫1 x(x + 2)
∫1 (
x x + 2)
Method 1: Partial Fractions
We first simplify by expressing the numerator in terms of the
denominator x 2 + 2x:
4 2
x2 + 4
x + 2x − 2x + 4
dx =
dx
∫1 x(x + 2)
∫1
x(x + 2)
=
∫1 (
1−
1
= (4 + 2 ln 4 − 4 ln 6) − (1 + 0 − 4 ln 3)
4
4
= [x + 2 ln | x | − 4 ln | x + 2 | ]
4
42 × 34
= 3 + ln
( 64 )
2x − 4
dx
x(x + 2) )
= 3 + ln 1
=3
We then decompose the integrand into partial fractions.
Let
A
B
2x − 4
= +
x(x + 2)
x x+2
∴ A(x + 2) + Bx = 2x − 4
letting x = 0: 2A = − 4 ⟹ A = − 2
letting x = − 2: −2B = − 8 ⟹ B = 4
Therefore
2x − 4
2
4
=− +
x(x + 2)
x x+2
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π
2
1
1
∫0 (4 − x 2) 32
cos x
98.
dx
∫0 5 − 3 sin x
99.
Method 1: Reverse Chain Rule
Method 1: Trigonometric Substitution
Observe that
d
(5 − 3 sin x) = − 3 cos x, hence we manipulate
dx
dx
We can make use of the Pythagorean identity 1 − sin2 θ = cos2 θ to
help simplify the integrand:
such that we can apply the reverse chain rule.
1
π
2
π
2
cos x
−3 cos x
1
dx = −
dx
∫0 5 − 3 sin x
3 ∫0 5 − 3 sin x
1
= − ln | 5 − 3 sin x |
[ 3
]
1
π
6
∫0 (4 − x 2) 2
dx =
3
2 cos θ
∫0 (4 − 4 sin2 θ) 23
π
6
π
2
=
2 cos θ
dθ
∫0 8 cos3 θ
0
π
6
=
=−
=
1
(ln 2 − ln 5)
3
1
sec2 θ dθ
∫
4 0
dθ
let x = 2 sin θ
dx
= 2 cos θ
dθ
d x = 2 cos θ dθ
when x = 0, θ = 0
π
when x = 1, θ =
6
π
1
6
= [tan θ]
4
0
5
1
ln
3
2
=
=
1
1
−0
4( 3
)
3
12
90
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π
2
100.
∫0
2 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ
Method 1: Double/Triple Angle Identity and Product to Sums
We simplify the integrand by recognising that:
sin(2θ) = 2 sin θ cos θ and sin(3θ) = 3 sin θ − 4 sin3 θ
π
2
∫0
2 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ =
π
2
∫0
sin 2θ sin 3θ dθ
Our integrand is now a product of two trigonometric functions, so
we apply the product to sum identities.
cos(A − B) − cos(A + B) = 2 sin A sin B
⟹ sin A sin B =
1
[cos(A − B) − cos(A + B)]
2
Letting A = 3θ and B = 2θ in our identity above:
π
π
2
1 2
2 sin θ cos θ(3 sin θ − 4 sin3 θ) dθ =
(cos θ − cos 5θ) dθ
∫0
2 ∫0
π
2
1
1
sin θ − sin 5θ
=
]
2[
5
0
=
1
1
1−
2(
5)
=
2
5
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