INTEGRATION
TECHNIQUES:
INTEGRATION… WHAT
DO WE KNOW SO FAR?
• In terms of getting integrals, we know how to do
basic anti-differentiation with power functions,
trig functions and exponential functions.
• We started some techniques using “usubstitution” to solve integrals. More or less, “a
reverse chain rule.”
• We also did area, volume, and other physical
applications using the definite integral.
• This chapter is devoted to the indefinite
integral.
1) U-SUBSTITUTION
REVISITED
• Just to freshen your minds,
let’s do a u-substitution
problem. There is a little
catch to it…
• For this problem, it is
preferable to pick u = x+2,
since du can never be in the
denominator.
• Use algebra to give a name
for x+1. Since x=u-2, then
x+1 would be u -2+1 or u-1.
• After putting everything in,
do the integration.
• Note that in the final
answer, the constant 2 does
not need to be there, since
C is more general.
x 1
 x  2 dx
u  x2
x u2
x 1  u  2 1  u 1
du  dx
u 1
1
 u du   1  u du
u  ln u
x  2  ln x  2
x  ln x  2  C
PRODUCT RULE
FLASHBACK
• Remember back in Chapter 2, we
mentioned the product rule? Looked like
the following…
In regular form….
Or differential form….
d (uv)
dv
du
u v
dx
dx
dx
d (uv)  udv  vdu
2) INTEGRATION BY
PARTS
• If you play around
with the differentials,
you will get the
following.
• If you integrate both
sides, you get what is
known as the
INTEGRATION BY
PARTS FORMULA
udv  d (uv)  vdu
udv

d
(
uv
)

vdu


 udv  uv   vdu
EXAMPLE#1
•
•
•
•
•
•
•
•
Integrate x*exdx.
STEP 1:
Pick your “u” and “dv.”
TRICK: L.I.P.E.T. This determines
what u should be initially. Logarithm,
Inverse, Power, Exponential, and Trig
functions.
In this problem, a power function and
an exponential function are present.
Since power (P) comes before
exponential (E), u should equal x.
From the u, find du. From the dv, find
v.
Therefore u=x, then du=dx! Don’t
forget that.
Simply plug this into the “parts
formula.”
 xe dx
x
ux
du  dx
ve
x
dv  e dx
x
EXAMPLE#1
• I = integral to be
solved.
• Plug in u, v, du, and dv.
• If the integral on the
right looks easy to
compute, then simply
integrate it.
• Don’t forget the +C!
I   udv  uv   vdu
I  xex   e x dx
I  xe  e  C
x
x
EXAMPLE#2
• Integrate the function…
e
sin
xdx

x
EXAMPLE#2
• Step 1: Find the u and
dv:
• An exponential and a
trig function is
present.
• E (exponential) comes
before T (trig).
• Use the exponential
function for u.
x
e
 sin xdx
ue
x
du  e dx
dv  sin xdx
v   cos x
x
EXAMPLE#2
• Step#2: Plug u, du, v and
dv into the formula
• Simplify…
• If the last integral is easy
to compute, compute it.
• However, it is not easy. In
fact, we are in more mess
than we started in. Looks
like we have to use the
integration by parts rule
again.
I   udv  uv   vdu
e
x
sin xdx  I
I  e x cos x   e x cos xdx
EXAMPLE#2
•
•
Use integration by parts again.
Since there is an exponential
function, call that “u.” After
finding u, du, v, and dv, plug those
values in the “Parts equation” and
see what happens…
•
We have even more of a
mess……but wait! The integral of
exsin(x) is supposed to be equal to
I (the integral we wanted to solve
for in the first place!!).
So we can replace the integral of
exsin(x) with I and add it to both
sides.
You will see that it works out.
•
Always add the constant 
•
I  e x cos x   e x cos xdx
u  ex
du  e x dx
v  sin x
dv  cos xdx


I  e x cos x  e x sin x   e x sin xdx
I   e x sin xdx
I  I  e x cos x  e x sin x  I  I
2 I  e x cos x  e x sin x
 e x cos x  e x sin x
I
C
2
RULE OF THUMB WITH
INTEGRATION BY PARTS
PROBLEMS
• Always pick the right “u.” If the problem is
getting really difficult, maybe you picked the
wrong “u.” Just like in u-substitution. You had to
pick the right “u” to work with the problem.
• If the integral on the right does not look easy to
compute, then do integration by parts for that
integral only.
• If you see the resulting integral looks like the
integral you are asked to solve for in the first
place, then simply combine the two like integrals
and use algebra to solve for the integral.
3) TRIGONOMETRIC
INTEGRALS
• You will always get the situation of funny
combination of integrals. Trig functions
are as such that you can translate from
one function to a function with just sines
and cosines. For example, you can always
write tan, sec, csc, cot in terms of either
sine or cosine.
• Remember the following identities from
PRE-CALCULUS!!!
IMPORTANT
IDENTITIES
sin 2 x  2 sin x cos x
cos 2 x  2 cos 2 x  1  1  2 sin 2 x  cos 2 x  sin 2 x
cos 2 x  sin 2 x  1
tan 2 x  1  sec 2 x
cot 2 x  1  csc 2 x
1
2
cos x  1  cos 2 x 
2
1
2
sin x  1  cos 2 x 
2
PROCEDURE
• Well… I’m afraid to say it, but there
is really no procedure or real
template in attacking these problems
except proper planning.
• This takes a great deal of practice
EXAMPLE# 1
• Given:
• Best thing to do is to
break the cosine
function down to a 2nd
degree multiplied by a
1st degree cosine.
• Since cos2x=1-sin2x,
you can replace it.
• Use u-substitution to
solve the integral.
3
cos
 xdx
 cos x cos xdx
 1  sin x cos xdx
2
2
u  sin x
du  cos xdx
2
1

u
 du
u3
sin 3 x
u   sin x 
C
3
3
EXAMPLE#2
• Given
• Break the 4th degree
sine to two 2nd degre
sines.
• Use the sin2x
theorem.
• Expand the binomial
squared.
• For the cos22x, use
the cos2x theorem.
• Simplify
• Don’t forget the +C!
4
sin
 xdx
 sin x sin x dx
2
2
1
 1  cos 2 x 
2


dx

1

cos
2
x
dx


 2 
4
1
2
1

2
cos
2
x

cos
2 x dx

4
1 
1 cos 4 x 
1

2
cos
2
x



dx
4 
2
2 
2


1 
3 cos 4 x 

2
cos
2
x



dx

4 
2
2 
3
sin 2 x sin 4 x
x

C
8
4
32
•
EXAMPLE #3
Whenever you see a
 cos
tan, sec, csc, or cot,
always convert them
to sines and cosines.
This way, you can
cancel or combine
whenever necessary.
• In this case,
tan2x=sin2x/cos2x. We
are also lucky that the
cos2x cancels.
• Using the sin2x
theorem, we can
simply integrate.
2
x tan 2 xdx
2

sin x 
2
 cos x cos 2 x dx
2
sin
 xdx
 1  cos 2 xdx 
sin 2 x
x
C
2
NOTE ABOUT
TRIGONOMETRIC
INTEGRALS
• There is no real rule for such integrals.
But always remember:
• 1) If there is a mix of sines and cosines,
break them up until they resemble an
easier form
• 2) Use any trig theorem that would be
relevant to make a problem simpler.
• 3) Convert everything to sines or cosines.
4) TRIGONOMETRIC
SUBSTITUTION
• Remember when we took derivatives of
inverse trigonometric functions, we commonly
dealt with sums or differences of squares.
• Similarly, integrating sums of differences of
square will lead us to the inverse trig
functions.
• However we need a stepping stone to
integrate such functions.
Given that a is a constant, and u is a function, then follow the
GENERAL PROCEDURE
a u
2
a u
2
u a
2
2
2
2
u  a sin 
u  a tan 
u  a sec 
THETA?
• Since we are working with a
substitution, theta would be the
variable to use subsitution…
• Doesn’t make sense? Let’s do an
example problem…
EXAMPLE #1
•
•
•
Initially a very bad looking
problem…
Focus on the denominator,
inside the radical, you have 9x2. In effect, that is a2-u2, a
being 3 and u being x. If a2-u2
is used, then according to the
table in the last slide, we
would use u=a sin.
You already found a name for
x. You need to give a name for
“dx.” Differentiate x with
respect to . Solve for dx.

x
2
9 x
ux
2
dx
a3
u  a sin 
x  3 sin 
dx  3 cosd
EXAMPLE #1
• Here is the original
problem…
• With the
substitutions of x and
dx, here is the original
problem…
• Simplify a little…
• Pull out constants
when needed.

x2
9 x
2
dx
3 sin  2 3 cosd
 9  3 sin  2


9 sin 2 
3 cosd
9  9 sin 
9 sin 2 
sin 2  cos
cosd  9
d
2
2
1  sin 
1  sin 
2
EXAMPLE #1
• The denominator is
actually cos2x,
according to the trig
theorem. (Memorize
them!)
• Simplify
• Integrate
• We have our answer in
terms of !!! We need
it in terms of x!
9
9
sin 2  cos 
1  sin 
sin 2  cos 
2
cos 2 
d
sin 2  cos 
d  9
d
cos 
9 sin 2 d  9 1  cos 2 d
sin 2 

9 

2


EXAMPLE #1
• Don’t forget what
we said earlier.
That x=3 sin . We
need to know what
 is in order to find
out the solution in
terms of x.
x  3 sin 
x
 sin 
3
 x
3
sin 2 x  cos 2 x  1
  arcsin  
x2
 cos 2 x  1
9
9  x2
 cos 2 x
9
1
9  x 2  cos x
3
sin 2  2 sin  cos  
2
 x  1
 2x 9  x
2  9  x 2  
9
 3  3

EXAMPLE #1
• Simply replace all
the q expressions
with x expressions.
• Simplify
• Add constant!!
• Sighs!! We’re
finished!!!
sin 2 

9 

2 

2 

 x x 9 x 

9 arcsin   



3
9




 x
9 arcsin    x 9  x 2  C
3
That was a LOT of work!!!
• Here are trig substitution steps…
• 1) Find the correct equality statement
using the table.
• 2) Make the proper substitutions.
Remember to have a substitute for x as
well as dx.
• 3) Integrate in terms of .
• 4) Convert all  terms to x terms.
5) RATIONAL
FUNCTIONS
• Of course, there will always be functions
in the form of a ratio of two functions.
• Two integrate most rational functions, the
method of partial fractions come into
play.
• <<Break from Calculus… entering Algebra
Territory>>
PARTIAL FRACTIONS
• This means you take a fraction and
break it down into a sum of many
fractions.
• This way, we can add up the integrals
of simpler easier fractions.
EXAMPLE
• Given
• The denominator could be
factored to (x+5)(x-2).
This way we could have new
denominators for the two
new fractions.
• Add these new fractions
and distribute. Make sure
you bring all x terms
together, as well as
bringing all the constants
together.
x 9
 x 2  3x  10 dx
B
A
x 9


x  5x  2 x  5 x  2
A x  2   B x  5
x  5x  2
Ax  2 A  Bx  5 B Ax  Bx  2 A  5 B

x  5x  2
x  5x  2
x A  B   2 A  5 B
x  5x  2
EXAMPLE
•
•
•
•
•
The coefficient of x on the
right side is 1. In order to
keep the equality true, the
coefficient of x on the left
side should also equal 1.
A+B=1
Same thing with the constant.
If the equality holds true,
then -2A+5B must equal -9.
To solve for A and B, you use
methods from algebra.
<System of linear equations>.
If you multiply A+B=1 by 2,
you will see that B=-1.
Therefore A=2.
x A  B   2 A  5B
x 9

x  5x  2
x  5x  2
A B 1
 2 A  5B  9
B  1
EXAMPLE
• Since A=2 and B=-1,
we can simply plug
them in.
• And integrate!!!
• And the final
answer!!
A
B

x5 x2
2
1
 x  5  x  2 dx
1
1
2
dx  
dx
x5
x2
2 ln x  5  ln x  2  C
ANOTHER EXAMPLE
• Given
• Note: If the
numerator has a
higher degree than
the denominator, then
do long polynomial
division.
• If you actually do the
long division, you will
get x-1-1/(x+1). This is
very easy to integrate.
x2
 x  1 dx
x2
1
 x  1 dx   x  1  x  1 dx
x2
 x  ln x  1  C
2
POINTERS OF PARTIAL
FRACTIONS
• 1) Check if the top degree is bigger
than the bottom. If so, perform long
division
• 2) If the denominator is factorable,
then assume that the denominators
of the new fractions will be those
factors.
6) QUADRATIC
DENOMINATOR
PROBLEMS
• This is really no different than
trigonometric substitution.
• Strictly rational functions with a quadratic
denominator that cannot be reduced.
• To make the denominator easier to work
with, you must complete the square
COMPLETING THE
SQUARE
•
•
•
•
•
•
•
•
Given problem:
Look at the denominator. Take
the coefficient of x and divide
it by 2.
Take this result and square it.
4/2 =22=4
This result would form a
perfect square when added to
x2+4x.
The perfect square would be
(x+2)2.
However, we have a 5. If you
add and subtract 4, combining
5 and -4 will yield 1.
You have a form of u2+a2! Time
for trig substitution!
3x  2
 x 2  4 x  5 dx
3x  2
 x 2  4 x  5   4  4 dx
3x  2
3x  2
 x 2  4 x  4  1 dx   x  22  1 dx
EXAMPLE
• Since we have
u2+a2, we must use
the fact of
u=a*tan().
• u=x+2 while a=1
• Substitute the
values in
appropriate spots.
3x  2
 x  22  1 dx
u  x2
a 1
u  a tan 
x  2  tan 
x  tan   2
dx  sec 2 d
3tan   2   2
 tan  2  1 dx
EXAMPLE (work)
3tan   2   2 2
 tan  2  1 sec d
3 tan   6  2 2
 tan 2   1 sec d
3 tan   8 2
 sec 2  sec d
tan 
 3 sec 2   8d
3
 2 sin 2  8d
3
 cos 2  8
4
x  2  tan 
tan 2   1  sec 2 
x  22  1  sec 2 
x 2  4 x  5  sec 2 
1
2

cos

2
x  4x  5
cos 2  2 cos 2   1
cos 2  2
1
1
2
x  4x  5
3
 cos 2  8
4
3
1

  2
  8 arctan  x  2   C
2  x  4x  5 
POINTERS
• 1) Make the denominator into one of
the three forms that allows trig
substitution by the use of completing
the square.
• 2) Follow rules of trig substitution.
7) IMPROPER
INTEGRALS
• This is not an integral evaluating
technique.
• An improper integral is basically an
integral that has infinity as its limits
or has a discontinuity within its
limits.
IMPROPER INTEGRALS
• Examples of improper integrals:


0
x
e dx
1
dx
3 x  2
2
IMPROPER INTEGRALS
• With limits of infinity, just
use a letter to replace the
infinity and treat as a
limit.
• And integrate as if nothing
ever happened 
• Don’t forget to use the
limit.
• Amazing! As we start from
0 to infinity, we get closer
to 1 square unit of area!
We say that it converges
to 1.


0


0
x
e dx  lim
b
e
x
b  0
x
e dx  lim

lim  e
b 


x b
0

b
b  0
dx
e  x dx

  lim e b  e 0

b 
 lim e  1  0  1  1
b 
b

•
•
•
IMPROPER INTEGRALS
Since we have a discontinuity in
this function at x=-2. To take this
into account, we must split the
integral into two parts. In addition,
we cannot go exactly -2, but we
have to get there pretty darn
close. Therefore, we must use the
one-sided limits from Chapter 1 to
represent this. Two integrals: one
from -3 to a little before -2, and a
little after -2 to 2.
Other than that, simply integrate lim
h 0

2h
2
1
1
1
dx

lim
dx

lim
dx
3 x  2


2

h
h 0 3
h

0
x2
x2
2
 2 h
2
1
1
dx  lim 
dx
h 0  2  h x  2
x2

lim ln x  2 
 lim ln x  2 
lim ln  2  h  2   ln  1 
 lim ln 2  2   ln  2  h  2 
3
 2 h
2
3
2 h
h 0
Notice how we got an answer that h0
 2 h
don’t exist!! D.N.E (does not exist)!
3
This means that this integral
h 0
h 0
diverges. Also if an integral goes to
ln 0  ln 1  ln 4  ln 0    0  ln 4    d .n.e.
infinity, it diverges.
2
 2 h
POINTERS OF
IMPROPER INTEGRALS
• Remember to identify all the points of
discontinuity. Remember to use limits
before and after the points of
discontinuity.
• If you have infinity as your limit,
remember to use infinity as your limit.
• Other than that, use ALL of the previous
techniques of integration mentioned.
FUNCTIONS WITHOUT
AN ANTIDERIVATIVE
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Besides three more chapters, this is the last of the single variable
calculus. That is to say: y=f(x) in 2-dimensional x,y graph.
Before moving on, I must admit… even though all continuous functions
have derivatives, not all continuous functions have simple integrals in
terms of elementary functions.
Elementary functions are adding, subtracting, multiplication, division,
power, rooting, exponential, logarithmic, trigonometric, all of their
inverses as well as combinations or composition functions. Basically, all
the functions you ever used were composed of elementary functions.
Some functions do not have elementary antiderivatives. For example…
the classic (sin x)/x problem.
No matter what method you tried. Neither by u-substitution, integration
by parts, trig substitution, partial fractions, or even guess and check will
get you an antiderivative.
From my experience from differential equations class last year, the
integral of sin(x)/x is Si(x) also known as the sine integral!
YOU DON’T NEED TO KNOW THAT!!!!
OUTTA THIS WORLD
FUNCTIONS!!!!!!!
• You will be dealing with functions like
erf(x), Si(x), Ci(x), Shi(x), Chi(x),
FresnelS(x), and FresnelC(x). Take
their derivatives and you’ll get
regular “sane” functions.  AAHHH!!!
HARI BOL!!!!
SUMMARY
• Actually, for once, looking at the length and material of this
chapter. I am quite amazed to say that I have no words to
summarize this chapter. There has been so many methods
of integration. Namely u-substitution, integration by parts,
how to deal with trig integrals, trig substitution, partial
fractions, quadratic denominators, and improper integrals.
• All I can say is that review this material over again!!!
• Like I said previously, there is no set way to do these
problems. There are more than one way of doing it.
• You have to know what to do when which problem arrives at
you.