techniques of integration.pdf

Techniques of Integration

In this chapter we will expand our toolkit of integration techniques. At this pointthe only technique, other than just recognizing an antiderivative, that we have de-veloped is u-substitution. By the time we are finished it will require some insightto choose the correct technique for each problem. Sometimes there will be morethan one method that works, though one technique may be much simpler thananother.

Remember that integration reverses differentiation. In particular u-substitutionreverses the chain rule for derivatives. We begin this chapter by exploiting a tech-nique that reverses another of the basic derivative rules, the product rule.

8.1 Integration by Parts: Reversing the Product Rule

Because integration reverses differentiation, every derivative rule can be ‘reversed’to become an integration rule. Sometimes this is easy; other times it is not. In thissection we will reverse the product rule for derivatives. Let’s set up the situation.

Suppose that u(x) and v(x) are both differentiable functions. Then the productrule says

ddx

[u(x)v(x)] = u(x)v0(x) + u0(x)v(x).

A short-hand way to write this is

ddx

(uv) = udvdx

+ vdudx

. (8.1)

Of course, if we integrate both sides of (8.1), we getZ d

dx(uv) dx =

Zu

dvdx

dx +Z

vdudx

dx.

On the left side of this equation integration undoes differentiation, and on theright side we can simplify the notation as we did in substitution problems so thatwe end up with

uv =Z

u dv +Z

v du.

Solving for the first term on the right side, we findZ

u dv = uv�Z

v du.

We have used the product rule to develop a new antidifferentiation rule. Thisresult is sufficiently important that we single it out as a theorem.

THEOREM 8.1 (Integration by Parts Formula). Suppose that u(x) and v(x) are both differen-tiable functions. Then Z

u dv = uv�Z

v du.

math 131, techniques of integration trig integrals 2

Take-Home Message. Here’s how to read Theorem 8.1: We can trade one inte-gral that we cannot do, say

Ru dv for another that we can (might be able) to do,

namely,R

v du. To do this we need to be able to identify a function u and thederivative of another function v in the original integrand. This technique is es-pecially useful when an integrand contains two functions that are not related toeach other. By the way, it is called ‘integration by parts’ because u and v are the‘parts.’ Let’s take a look at some examples.

EXAMPLE 8.1.1. Consider the integralZ

x cos(x) dx.

The function x is unrelated to the function cos(x). Compare this toZ

x cos(x2) dx.

In this second situation, the function x is related to cos(x2). In particular, x is almost thederivative of the ‘inside function’ x2—we are off only by a constant. You should recognizethis second problem as a u-substitution problem. But what about the first integral?

SOLUTION. Let’s see how to use integration by parts to solve the integral. In theoriginal integral

Rx cos(x) dx we have to identify a function u and the derivative of

another function v. Let’s try

u = x and dv = cos x dx

The integration by parts formula then saysR

u dv = uv �R

v du. So we need todetermine v and du from the information we have so far. Getting du is easy; just takethe derivative of u:

u = x ) du = 1dx = dx.

Getting v requires us to integrate dv:

v =Z

dv =Z

cos x dx = sin x.

Notice that we have not added +c to the solution for v. As we’ll see below, it turnsout not to matter in this situation. Thus uv = x sin(x) and

Rv du =

Rsin(x) dx. Notice

that we are able to do this latter integral! We have exchanged the integral we couldnot determine,

Rx cos(x) dx, for one that we can,

Rsin(x) dx.

Substituting all of this into the integration by parts formula producesZ

u dv = uv�Z

v du )Z

x cos(x) dx = x sin(x)�Z

sin(x) dx

= x sin(x) + cos(x) + c.

We can check that this is correct by differentiating our answer:

ddx

(x sin(x) + cos(x) + c) = sin(x) + x cos(x)� sin(x) = x cos(x),

which is the original integrand. Notice how the product rule comes into play inchecking the answer.

‘What if’ #1. What if we had chosen ‘the parts’ differently. Suppose we let u = cos xand dv = x dx. Then

u = cos x ) du = � sin x dx.

and

v =Z

dv =Z

x dx =x2

2.

Substituting all of this into the integration by parts formula producesZ

u dv = uv�Z

v du )Z

x cos(x) dx =x2

2cos(x)�

Z� x2

2sin(x) dx


which we still do not know how to integrate. In fact, if anything, the new integral is‘worse’ than the original one because of the x2-term. The power of x has increased. Inour original solution, the power of x decreased because we let u be the function x. Itsderivative du = dx appeared in the new integrand (as well as v), so the new integralwas simpler. We will list a few guidelines below that may help in selecting the ‘parts.’

‘What if’ #2. What if we had added a constant +C to v in the original solution (so thatv + C = sin x + C). Let’s see what happens in the general integration by parts formulawhen we substitute v + c for v.

Zu dv = u(v + C)�

Z(v + C) du = uv + Cu�

Zv du�

ZC du

= uv + Cu�Z

v du� Cu

= uv�Z

v du.

We get the same answer as in the original integration by parts formula because Cuand �Cu end up canceling each other. In most cases it is simpler to just ignore the+C.

EXAMPLE 8.1.2 (Classic Parts). DetermineZ

xe�x dx.

SOLUTION. We have unrelated functions x and e�x so integration by parts may beuseful, since our other technique of substitution does not make sense here. We presentthe solution in tabular form which shows the parts in the left column and the integra-tion in the right column.

u = x dv = e�x dxR

u dv = uv�R

v du

du = dx v =R

dv =R�e�x dx = e�x R

xe�x dx = �xe�x �R�e�x dx

= �xe�x � e�x + c


ddx

��xe�x � e�x + c

�= �e�x + xe�x + e�x = xe�x,

which is the original integrand.

EXAMPLE 8.1.3. DetermineZ

ln x dx.

SOLUTION. Careful, we don’t want the derivative of ln x, we want the antiderivative.This time we have no choice. Since we don’t know how to integrate ln x yet, we haveto differentiate it. In other words we want u = ln x and that leaves dv = dx. So

u = ln x dv = dxR

u dv = uv�R

v du

du = 1x dx v =

Rdv =

Rdx = x

Rln x dx = x ln x�

R1 dx = x ln x� x + c


ddx

(x ln x� x + c) = ln x + x✓

1x

◆� 1 = ln x,


Some Tips for Using Integration by Parts

First, remember that this technique is most useful when there are unrelated func-tions in the integrand. Second, right now you have only two techniques: substi-tution and parts. Substitution is usually easier to carry out, so first check to see


if that technique applies. At this point, if it does not, then parts is likely to be themethod that works. Third, if you choose parts and the integral gets ‘worse’ af-ter the first substitution, step back and ask yourself whether the parts should bechosen differently. Be prepared to change; you may not always choose the partscorrectly the first time.

Here are a couple of general principles that will help you select u and dv whenusing integration by parts. There really is no substitute for practice, but these mayhelp you be more efficient.

• Try selecting dv as the most complicated portion of the integrand that you canintegrate. . . and let u be the rest of the integrand.

• Try letting u be the portion of the integrand whose derivative is simpler thanu, e.g., du might end up being a lower power of x. This is what we did in Ex-ample 8.1.1 when we selected u = x. There du = 1 dx was simpler. If we hadchosen u = cos x, then du = � sin x dx would just as complicated as u. The newintegrand would not have been simplified.

Typical Scenarios for Integration by Parts

The next few examples illustrate several basic situations where integration by partsis useful.

EXAMPLE 8.1.4 (Unrelated Parts). DetermineZ p

x ln x dx.

SOLUTION. The two functions in the integrand,p

x and ln x are not related to eachother. Parts should immediately come to mind. Since ln x is not simple to integrate(though we know how to do it from Example 8.1.3), it makes sense to use it as u. So

u = ln x dv =p

x dxR

u dv = uv�R

v du

du = 1x dx v =

Rdv =

R px dx = 2

3 x3/2 R px ln x dx = 2

3 x3/2 ln x�R 2

3 x3/2 · 1x dx

= 23 x3/2 ln x�

R 23 x1/2 dx

= 23 x3/2 ln x� 4

9 x3/2 + c

You should check that this is correct by differentiating the answer.

math 131, techniques of integration integration by parts, part ii 5

8.2 Doing It Twice: One Good Turn Deserves Another.

There are a couple of situations where using integration by parts twice is just theticket. Part way through you should be able to see why this works.

EXAMPLE 8.2.1 (Parts: Reduction). DetermineZ

x2 sin x dx.

SOLUTION. The two functions in the integrand, x2 and sin x are unrelated to eachother. Parts should come to mind. Since sin x is the more complicated piece and canbe integrated, use it as dv. Likewise, x2 becomes simpler when integrated (sin x doesnot), so use it as u. So (being careful with signs)

u = x2 dv = sin x dxR

u dv = uv �R

v du

du = 2x dx v =R

dv =R

sin x dx = � cos xR

x2 sin x dx = �x2 cos x +R

2x cos x dx

Notice that the new integralR

2x cos x dx is not immediately ‘doable’ but is simplerthan the original one (the power of x is lower) and can be done with parts.

u = 2x dv = cos x dxR

x2 sin x dx = �x2 cos x +R

2x cos x dx

du = 2 dx v =R

dv =R

cos x dx = sin x = �x2 cos x + 2x sin x �R

2 sin x dx

= �x2 cos x + 2x sin x + 2 cos x + c

Check that this is correct by differentiating the answer.

ddx

⇣�x2 cos x + 2x sin x + 2 cos x + c

⌘= �2x cos x + x2 sin x + 2 sin x + 2x cos x � 2 sin x

= x2 sin x,


EXAMPLE 8.2.2 (Parts: Circular Reasoning). DetermineZ

ex sin x dx.

SOLUTION. The two functions in the integrand, ex and sin x are unrelated to eachother. Think parts! Here it really does not matter which we use as u or dv, though Iwould give a slight preference to using u = sin x since it is slightly easier to differenti-ate sin x than it is to integrate it (where we get a negative sign). So just to be contraryI will do it the other way. We will use parts twice, as it turns out.

u = ex dv = sin x dxR

ex sin x dx = �ex cos x +R

ex cos x dxdu = exdx v = � cos x Watch the signs! This is what I meant above.

Notice that the new integralR

ex cos x dx is not a whole lot different than the first one.Parts is indicated. Important: Continue to set up the parts in the same way: u is stillthe exponential function and dv is still the trig function.

u = ex dv = cos x dxR

ex sin x dx = �ex cos x + ex sin x �R

ex sin x dx. Signs!

du = exdx v = sin xR

ex sin x dx appears on both sides of the equation.

Solve for it: 2R

ex sin x dx = ex(sin x � cos x) + c

Thus,R

ex sin x dx = 12 ex(sin x � cos x) + c

You should check that this is correct by differentiating the answer. Notice what hap-pened here. We never actually did one of the integrals. We kept applying the partsformula until the problem more or less circled around on itself. In these sorts of prob-lems you must be very careful with the signs attached to the various integrals. Theseare great problems to demonstrate your mastering of this technique.


EXAMPLE 8.2.3 (Udoable Becomes Doable). DetermineZ

arcsin x dx. Sometimes we simplydon’t know an antiderivative for a familiar function. Parts can be a way to solve the prob-lem (see Example 8.1.3 for another example of this).

SOLUTION. We don’t have much choice. If parts applies we must let u = arcsin xsince we do not know how to integrate it. So

u = arcsin x dv = dxR

arcsin x dx = x arcsin x �R xp

1�x2 dx

du = 1p1�x2 dx v =

Rdv =

Rdx = x Substitution u = 1 � x2 so du = �2x dx ) 1

2 du = �x dxR

arcsin x dx = x arcsin x +R 1

2 u�1/2 du Signs!!R

arcsin x dx = x arcsin x +p

u + c = x arcsin x +p

1 � x2 + c

You should check that this is correct by differentiating the answer. Another greatproblems to demonstrate your understanding of integration techniques.

EXAMPLE 8.2.4 (Tricky). DetermineZ xe2x

(2x + 1)2 dx.

SOLUTION. It does not look like a substitution problem, so it is probably a partsproblem. Here the choice of parts is tricky. But if we keep in mind the suggestion thatwe use the “most complicated portion of the integrand that you can integrate for dv",then let’s use dv = 1

(2x+1)2 dx. Then by the power rule and a ‘mental adjustment’,

v =Z(2x + 1)�2 dx = �1

2(2x + 1)�1 = � 1

2(2x + 1).

Now u is the rest of the integrand so u = xe2x and

du = (2xe2x + e2x dx = (2x + 1)e2x dx.

Note the factor of 2x + 1. Putting this all together (and keeping track of signs),Z xe2x

(2x + 1)2 dx = uv �Z

v du = � xe2x

2(2x + 1)+

Z(2x + 1)e2x

2(2x + 1)dx

= � xe2x

2(2x + 1)+

Z e2x

2dx

= � xe2x

2(2x + 1)+

e2x

4+ c

where we used a ‘mental adjustment’ at the last step. You should check that this iscorrect by differentiating the answer.

8.3 Problems

1. Integral Mix Up: Before working these out, go through and classify each by the tech-nique that you think will apply: substitution, parts, parts twice, or ordinary methods.Which can’t you do yet? The answers are below.

(a)Z

2e�px dx (b)Z

cos xesin x dx (c)Z

ex cos x dx

(d)Z

x cos x dx (e)Z

cos(2px) dx (f )Z ln x

xdx

(g)Z(x2 + 1)ex3+3x dx (h)

Z(x2 + 1)ex dx (i)

Zx2 ln x dx

(j)Z

sec2(2x) dx (k)Z x

25 + x2 dx (l)Z 1

1 + 25x2 dx

(m)Z 1p

1 � 9x2dx (n)

Z cos xp1 � sin2 x

dx (o)Z sin�1 xp

1 � x2dx


1. Some Answers to the Mix-Up Problem: (All “+c".)

(a)�2e�px

p(b) esin x (c) 1

2 ex(cos x + sin x)

(d) x sin x + cos x (e) 12p sin(2px) (f )

(ln x)2

2(g) 1

3 ex3+3x (h) (x2 � 2x + 3)ex (i) 13 x3

⇣ln x � 1

3

⌘

(j) 12 tan(2x) (k) 1

2 ln(25 + x2) (l) 15 arctan 5x

(m) 13 arcsin 3x (n) arcsin(sin x) (o) 1

2 (arcsin x)2

Parts: Further Examples

We end with a few more examples that involve trig functions and that provide asegue to our foray into trig integrals more generally.

EXAMPLE 8.3.1 (Oddball). DetermineZ

cos(ln x) dx. Careful! This is not a product of func-tions, it is a composition.

SOLUTION. Again we don’t have much choice. It’s not substitution since the argu-ment1 is u = ln x and du is nowhere in sight. So let’s try parts. Since the integrand 1 The term ‘argument’ means the input

to the functionconsists of a single function that we don’t know how to antidifferentiate we must letu = cos(ln x). So

u = cos(ln x) dv = dxR

cos(ln x) dx = x cos(ln x) +R

x · 1x sin(ln x) dx Sign!

du = � 1x sin(ln x) dx v = x = x cos(ln x) +

Rsin(ln x) dx

The new integralR

sin(ln x) dx looks much like the old one, so we try parts again andhope that we can ‘circle around’ to where we started as in Example 8.2.2.

u = sin(ln x) dv = dxR

sin(ln x) dx = x sin(ln x)�R

x · 1x cos(ln x) dx

du = 1x cos(ln x) dx v = x = x sin(ln x)�

Rcos(ln x) dx Cycles back

Putting the two parts integrals together:Z

cos(ln x) dx = x cos(ln x) + x sin(ln x)�Z

cos(ln x) dx

which gives

2Z

cos(ln x) dx = x cos(ln x) + x sin(ln x)

or Zcos(ln x) dx =

12(x cos(ln x) + x sin(ln x)) + c

You should check that this is correct by differentiating the answer.

EXAMPLE 8.3.2 (Trig). DetermineZ

sin(x) cos(4x) dx. Careful! How does this problem differ

fromZ

sin(4x) cos(4x) dx? The latter we can do by substitution.

SOLUTION. It’s not substitution, so we try parts. It is a little easier to integrate sin xrather than cos(4x) because of the constants involved. So

u = cos(4x) dv = sin x, dxR

sin(x) cos(4x) dx = � cos(x) cos(4x)�R

cos(x)4 sin(4x) dx

du = �4 sin(4x) dx v = � cos x Parts again! Should circle around.

The new integral �R

cos(x)4 sin(4x) dx looks much like the old one, so we try partsagain and hope that we can ‘circle around’ to where we started as in Example 8.2.2.


Remember to choose the parts in the same way at each stage.

u = 4 sin(4x) dv = cos x dx

du = 16 cos(4x) dx v = sin x

SoZ

sin(x) cos(4x) dx = � cos(x) cos(4x)�Z

cos(x)4 sin(4x) dx

= � cos(x) cos(4x)�

4 sin(x) sin(4x)� 16Z

sin(x) cos(4x) dx�

.

This gives

�15Z

sin(x) cos(4x) dx = � cos(x) cos(4x)� 4 sin(x) sin(4x)

or Zsin(x) cos(4x) dx =

cos(x) cos(4x)� 4 sin(x) sin(4x)15

+ c

Not too bad, but be careful of your signs throughout similar problems. You shouldcheck that this is correct by differentiating the answer.

Integrating Low Powers of the Secant Function. Remember that we were able to solveRsec x dx using substitution. We saw

Zsec x dx = ln | sec x + tan x|+ c.

Of course Zsec2 x dx = tan x + c.

EXAMPLE 8.3.3 (Trig). What about Zsec3 x dx?

SOLUTION. Following the suggestions for integration by parts, the most complicatedfactor in the integrand that we can integrate is dv = sec2 x dx. Working this out we get

u = sec x dv = sec2 x dxdu = sec x tan x dx v = tan x

SoZ

sec3 x dx =Z

sec x sec2 x dx = sec x tan x �Z

sec x tan2 x dx

= sec x tan x �Z

sec x(sec2 x � 1) dx

= sec x tan x �Z

sec3 x � sec x dx

Solving for the original integral we get

2Z

sec3 x dx = sec x tan x +Z

sec x dx = sec x tan x + ln | sec x + tan x|.

SoZ

sec3 x dx =sec x tan x + ln | sec x + tan x|

2+ c.

We will look at some additional trig integrals in the next section.


8.4 Trig Integrals

This section is devoted to integrating powers of trig functions. First we examinepowers of sine and cosine functions.

Powers of a Single Trig Function

We begin with four key trig identities that you should memorize that will makeyour life and these integrals much simpler.

Four Key Identities.

cos2 u + sin2 u = 1 (so sin2 u = 1 � cos2 u or cos2 u = 1 � sin2 u)

1 + tan2 u = sec2 u (so tan2 u = sec2 u � 1)

sin2 u = 12 � 1

2 cos 2u (Half angle formula).

cos2 u = 12 + 1

2 cos 2u

The half angle formulas are used to integrate sin2 u or cos2 u in the obvious way.


cos2(8x) dx.

SOLUTION. Use equation (3) above with u = 8x. Note the use of a ‘mental adjust-ment.’

Zcos2(8x) dx =

Z12 + 1

2 cos(16x) dx = 12 x + 1

32 sin(16x) + c.

We have already seen how to integrate low powers of the secant and tangentfunctions.

Rtan u du = ln | sec u|+ c

Rtan2 u du =

Rsec2 u � 1 du = tan u � u + c (Use a key trig identity.)

Rsec u du = ln | sec u + tan u|+ c

Rsec2 u du = tan u + c

Higher Powers of Trig Functions

When we want to integrate higher powers (powers greater than n � 2) of a singletrig function we can make use of integration by parts.

EXAMPLE 8.4.2. Suppose that n � 2 is an integer. DetermineZ

sinn x dx.

SOLUTION. Use integration by parts. The key is to write sinn x as sinn�1 x sin x. u = sinn�1 x dv = sin x dxdu = (n � 1) sinn�2 x cos x dx v = � cos xZ

sinn x dx = � sinn�1 x cos x + (n � 1)Z

sinn�2 x cos2 x dx

= � sinn�1 x cos x + (n � 1)Z

sinn�2 x(1 � sin2 x) dx

= � sinn�1 x cos x + (n � 1)Z

sinn�2 x � sinn x dx�

Combining all the sinn x terms,

nZ

sinn x dx = � sinn�1 x cos x + (n � 1)Z

sinn�2 x dx


So we obtainZ

sinn x dx = � sinn�1 x cos xn

+n � 1

n

Zsinn�2 x dx.

This is what is known as a reduction formula, and it can be used repeatedly to deter-mine integrals of high powers of sin x.


sin5 x dx.

SOLUTION. Use the reduction formula with n = 5.Z

sin5 x dx = � sin4 x cos x5

+45

Zsin3 x dx.

Now use it again with n = 3.Z


+45

"� sin2 x cos x

3+

23

Zsin x dx

#.

Now we can finish the integration in the usual way.Z


� 4 sin2 x cos x15

� 8 cos x15

+ c.

Reduction Formulas for Large Powers. There are reduction formulas for the othertrig functions as well. They are verified using integration by parts. The most im-portant are (these do not need to be memorized). Repeated application may benecessary.

Rcosn u du = 1

n cosn�1 u sin u + n�1n

Rcosn�2 u du

Rsinn u du = � 1

n sinn�1 u cos u + n�1n

Rsinn�2 u du

Rtann u du = 1

n�1 tann�1 u �R

tann�2 u duR

secn u du = 1n�1 secn�2 u tan u + n�2

n�1R

secn�2 u du


tan4 x dx.

SOLUTION. Use the reduction formula with n = 4.Z

tan4 x dx =tan3 x

3�

Ztan2 x dx

=tan3 x

3�

tan x

1�

Z1 dx

�

=tan3 x

3� tan x + x + c.

Notice that we used the fact that tan0 x = 1.

Integrating Products of Powers of Sines and Cosines

We will now develop some guidelines for integrals of the formZ

sinm x cosn x dx

where either m or n is a positive integer. Notice that both trig functions have thesame argument. The goal is to use the power rule, as in the following: DetermineR

sin x cos4 x dx. You should recognize this as a simple substitution problem; letu = cos x and du = � sin x dx. Then

Zsin x cos4 x dx = �

Zu4 du = �u5

5+ c = �cos5 x

5+ c.

Thus, the protocol is to use trig identities and u-substitution to turn the integralinto a simple power rule problem.


Guidelines for Products of Powers of Sines and Cosines: These general principles canhelp you solve integrals of the form

Rsinm x cosn x dx.

1. If the power of sine is odd and positive, split off a factor of sine for du andconvert the rest to cosines, let u = cos x, and then integrate. For example,

Zm=2k+1 oddz }| {sin2k+1 x cosn x dx =

Z(sin2 x)kx cosn x ·

use for duz }| {sin x dx

=Z

convert to cosinesz }| {(1 � cos2 x)k cosn x · sin x dx = �

Z(1 � u2)kun du.

2. If the power of cosine is odd and positive (and the power of sine is even), splitoff a factor of cosine for du and convert the rest to sines, let u = sin x, and thenintegrate. For example,

Zsinm x

n=2k+1 oddz }| {cos2k+1 x dx =

Zsinm x(cos2 x)k ·

use for duz }| {cos x dx

=Z

sinm x

convert to sinesz }| {(1 � sin2 x)k · cos x dx =

Zum(1 � u2)k du.

3. If both powers of sine and cosine are even and non-negative, make repeated useof the identities sin2 u = 1

2 � 12 cos 2u and cos2 u = 1

2 + 12 cos 2u to powers of

cosines. Then use reduction formula #1.

4. Use a table of integrals or WolframAlpha or other software. Certainly you shoulduse this tool in later courses whether in math or other departments.

EXAMPLE 8.4.5 (Using Rule 1). DetermineZ

sin3 x cos2 x dx.

SOLUTION. Since the power of the sine function is odd, we use Guideline #1.Z

sin3 x cos2 x dx =Z

sin2 x cos2 x · sin x dx split off a power of sin x

=Z(1 � cos2 x) cos2 x · sin x dx use a trig id

= �Z(1 � u2)u2 du substitute u = cos x, du = � sin x

= �Z

u2 + u4 du expand

= �u3

3+

u5

5+ c

= � cos3 x3

+cos5 x

5+ c


sin4 3x cos3 3x dx.

SOLUTION. Both functions have the same argument (angle). The sine function has an


even power and the power of the cosine function is odd; we use rule #2.Z

sin4 3x cos3 3x dx =Z

sin4 3x cos2 3x · cos 3x dx split off a power of cos 3x

=Z

sin4 3x(1 � sin2 3x) cos 3x dx use a trig id

=13

Zu4(1 � u2) du u = sin 3x, 1

3 du = cos 3x dx

=13

Zu4 � u6 du expand

=u5

15� u7

21+ c

=sin5 3x

15� sin7 3x

21+ c

Gosh, be careful of the mental adjustment required for the substitution.


sin2 5x cos2 5x dx.

SOLUTION. The powers of the sine and cosine function are even, we use Guideline #3.Z

sin2 5x cos2 5x dx =Z ⇣

12 � 1

2 cos 10x⌘ ⇣

12 + 1

2 cos 10x⌘

dx half angle formula

=Z

14 � 1

4 cos2 10x dx expand

=Z

14 � 1

4

h12 + 1

2 cos 20xi

dx half angle formula

=Z

18 � 1

8 cos 20x dx combine terms

= 18 x � 1

160 sin 20x + c mental adjustment

Again not too bad! Just be careful.


cos5 x dx.

SOLUTION. Since the sine function does not appear and the power of the cosinefunction is odd, we can use Guideline #2. A preferred (?) method might be to use areduction formula.

Zcos5 x dx =

Z(cos2 x)2 cos x dx split off a power of cos x

=Z(1 � sin2 x)2 cos x dx use a trig id

=Z(1 � u2)2 du u = sin x, du = cos x dx

=Z

1 � 2u2 + u4 du expand

= u � 2u3

3+

u5

5+ c

= sin x � 2 sin3 x3

+sin5 x

5+ c

Not too bad!

Integrating Products of Powers of Secant and Tangent

Here’s some material from your text that we will use to integrate powers of tan-gent and secant. You do not need to memorize this. But you do need to know howto use this information. The idea is very similar to what we used for products ofsines and cosines.


Guidelines for Products of Powers of Tangents and Secants: These general principlescan help you solve integrals of the form

Rtanm x secn x dx.

1. If the power of secant is even and positive, split off sec2 x to use for du andconvert the rest to tangents, then let u = tan x, and integrate. For example,

Ztanm x

n=2k evenz }| {sec2k x dx =

Ztanm x(sec2 x)k�1 ·

use for duz }| {sec2 x dx

=Z

tanm x

convert to tangentsz }| {(1 + tan2 x)k�1 · sec2 x dx =

Zum(1 + u2)k�1 du.

2. If the power of tangent is odd and positive (and the power of secant is odd),split off sec x tan x for du and convert the rest to secants, let u = sec x, and thenintegrate. For example,

Zm=2k+1 oddz }| {tan2k+1 x secn x dx =

Z(tan2 x)k secn�1 x ·

use for duz }| {sec x tan x dx

=Z

convert to secantsz }| {(sec2 x � 1)k secn�1 x · sec x tan x dx =

Z(u2 � 1)kun�1 du.

3. If m is even and n is odd, convert the tangents to secants and use the reductionformula for powers of the secant:

Zm=2k evenz }| {tan2k x

n oddz }| {secn x dx =

Zconvert to secantsz }| {(sec2 x � 1)k secn x dx.

4. In real life, use WolframAlpha, or look in a table of integrals.

EXAMPLE 8.4.9 (Using Guideline 2). DetermineZ

tan3 x sec3 x dx.

SOLUTION. The powers of both secant and tangent are odd, so Guideline #2 aboveapplies.

Ztan3 x sec3 x dx =

Ztan2 x sec2 x(sec x tan x) dx split off a secant-tangent

=Z(sec2 x � 1) sec2 x(sec x tan x) dx convert to secants

=Z(u2 � 1)u2 du u = sec x, du = sec x tan x

=Z

u4 � u2 du expand

=u5

5� u3

3+ c

= 15 sec5 x � 1

3 sec3 x + c


tan4 x sec4 x dx.


SOLUTION. The powers of both secant and tangent are even, so Guideline #1 applies.Z

tan4 x sec4 x dx =Z

tan4 x sec2 x · sec2 x dx split off a secant-squared

=Z

tan4 x(1 + tan2 x) · sec2 x dx convert to tangents

=Z

u4(1 + u2) du u = tan x, du = sec2 x

=Z

u4 + u6 du expand

=u5

5+

u7

7+ c

= 15 tan5 x + 1

7 tan7 x + c


tan2 x sec x dx.

SOLUTION. The powers of tangent is even and the power of secant is odd, so Guide-line #3 applies.Z

tan2 x sec x dx =Z(sec2 x � 1) sec x dx convert to secants

=Z

sec3 x dx �Z

sec x dx expand

=

sec x tan x

2+

12

Zsec x dx

��

Zsec x dx reduction formula

=sec x tan x

2� 1

2

Zsec x dx combine, simplify

=sec x tan x

2� 1

2ln | sec x + tan x|+ c.

Take-Home Message. OK, the goal here is to be able to use these guidelines forproducts of trig functions. Here’s what you must be able to do without lookingback to the guidelines:

1. You need to KNOW the four key identities.

2. You need to be able to integrate low powers (n = 1 or 2) of the four main trigfunctions.

3. You need to be able to integrate products of the form sinn x cosm x when atleast one of the powers is an odd positive integer by splitting off an appropriatefactor, using a key trig id, and then using u-substitution (Guidelines 1 and 2 forProducts of Sines and Cosines).

4. You do NOT need to memorize the reduction formulæ, do not need to memo-rize the guidelines for products of the form tanm x secn x.

Integration by Triangle Substitutions

The Area of a Circle

So far we have used the technique of u-substitution (i.e., reversing the chain rule)and integration by parts (reversing the product rule) to extend the “list" of func-tions that we can antidifferentiate. Remember that we are in the antidifferentiation‘business’ because the Fundamental Theorem (FTC) says that if F is an antideriva-tive of f , then the area under f on the interval [a, b] is

∫ ba f (x) dx = F(b) − F(a).

This theorem “solves" the area problem, at least for those functions whose an-tiderivatives we know. But even with our new integration techniques there aremany integrals we cannot yet do, such as:

EXAMPLE 7.2.1. Find the area of the semi-circle of radius r. Since middle school we havebeen told that the area of a circle is πr2, so the area of a semi-circle is 1

2 πr2. But why isthis formula valid? Recall that the entire circle of radius r centered at the origin is the setof points that satisfies x2 + y2 = r2. It follows that the upper semi-circle is given by thefunction y = f (x) =

√r2 − x2 on the interval [−r, r]. So the area we seek is∫ r

−r

√r2 − x2 dx.

It’s scandalous but true: none of the techniques we’ve developed so far will help us findan appropriate antiderivative. Still

√r2 − x2 should remind us of right triangles, and we’ve

excised one such triangle from the figure below.

−r 0 r

........

........

....................................................................................................................

..............................................................................................................................................................................................................................

f (x) =√

r2 − x2

........

.....

........

.....

........

.....

........

..................

..........................

....................... .................................................................

........

........

........

........

........

........

........

........

........

........

........

...................................................................................................................x

r

θ

√r2 − x2

Figure 7.1: Left: A semi-circle withequation y =

√r2 − x2. Right: A

corresponding right triangle.

The sides of the triangle are related to each other via the trigonometric functions ofthe angle θ. We can use these to set up a fancy u-substitution, or more appropriately, aθ-substitution.

√r2 − x2

r= sin θ ⇒

√r2 − x2 = r sin θ

xr

= cos θ ⇒ x = r cos θ

⇒ dx = −r sin θ dθ

Notice that each part of the original integrand can now be written in terms of the angle θ.Of course, we need to write the x-limits of integration as θ-limits. When

x = −r = r cos θ ⇒ cos θ = −1 ⇒ θ = π

x = r = r cos θ ⇒ cos θ = 1 ⇒ θ = 0

This makes sense: θ changes from π to 0 as you move around the semi-circle clockwise

math 131, techniques of integration iv triangle substitutions 2

from −r to r. Now the integral can be rewritten in terms of θ and solved.∫ r

−r

√r2 − x2 dx =

∫ 0

πr sin θ(−r sin θ) dθ = −r2

∫ 0

πsin2 θ dθ = −r2

∫ 0

π

12 −

12 cos(2θ) dθ

= −r2(

12 θ − 1

4 sin(2θ)) ∣∣∣0

π

= −r2[(0− 0)−

(π

2− 0)]

=πr2

2

From this it follows that the area of a circle of radius r is πr2.

7.3 General Triangle Substitutions

In general, triangle substitutions are based on, well, triangles! Right triangles, in particular.Three different substitutions arise depending on the integrand and how the sides of thetriangle are labeled. The triangles are based on the corresponding forms of the square rootsthat arise from the Pythagorean theorem in these triangles. The three cases are:

√x2 + a2,√

x2 − a2, and√

a2 − x2.

Case 1: Integrals involving√

x2 + a2

In this case,√

a2 + x2 must correspond to the hypotenuse of the right triangle. (Why?)So with the legs of the triangle labeled as below, we have:

....................................................................................................................................................................................................................................................................................................................................................................

a

√x2 + a2

θ

x

x = a tan θ

dx = a sec2 θ dθ√

x2 + a2 = a sec θ

EXAMPLE 7.3.1. Calculate the following indefinite integral:∫ 1

x2√

x2 + 4dx.

Using the triangle substitution in the box above with a2 = 4 or a = 2, we have:

∫ 1

x2√

x2 + 4dx =

∫ 14 tan2 θ · 2 sec θ

· 2 sec2 θ dθ = 14

∫ sec θ

tan2 θdθ = 1

4

∫ 1cos θ

sin2 θcos2 θ

dθ

= 14

∫ cos θ

sin2 θdθ

= 14

∫(sin θ)−2 cos θ dθ

= − 14 (sin θ)−1 + c

We must convert our answer back to a function of x. Look back at the triangle above. Noticethat sin θ = x√

x2+4. So

− 14 (sin θ)−1 + c = − 1

4

(x√

x2 + 4

)−1= −√

x2 + 44x

+ c.

One question you may have is how should you select which leg of the triangle shouldcorrespond to x and which to a? The answer is that the substitution will work no matterwhich you use, but it will be easier with x as the opposite side to angle θ. If we had let x bethe adjacent side, then x = a cot θ, a less familiar trig function.

EXAMPLE 7.3.2. Calculate the indefinite integral:∫ 1√

16 + x2dx.


Using the triangle substitution above with a2 = 16 or a = 4, we have:∫ 1√16 + x2

dx =∫ 4 sec2 θ

4 sec θdθ =

∫sec θ dθ = ln | sec θ + tan θ|+ c = ln |

√16+x2

4 + x4 |+ c


x2 − a2

In this case, x must correspond to the hypotenuse of the right triangle. (Why?) Againwe have our choice of how to label the legs, one side a and the other

√x2 − a2. With the

selection made below, x = a sec θ. What would x equal if we had let a be the side oppositeθ?

....................................................................................................................................................................................................................................................................................................................................................................

a

x

θ

√x2 − a2

x = a sec θ

dx = a sec θ tan θ dθ√

x2 − a2 = a tan θ

EXAMPLE 7.3.3. Evaluate the definite integral:∫ 2

1

√x2 − 1

xdx.

Use the triangle substitution above with a2 = 1 or a = 1. We also have to change limits.

x = 1 = sec θ ⇒ θ = 0x = 2 = sec θ ⇒ cos θ = 1

2 ⇒ θ = π/3

Now proceed with the substitution:

∫ 2

1

√x2 − 1

xdx =

∫ π/3

0

tan θ

sec θ· sec θ tan θ dθ =

∫ π/3

0tan2 θ dθ =

∫ π/3

0sec2 θ − 1 dθ

= tan θ − θ∣∣∣π/3

0

=√

3− π/3


a2 − x2

This was situation in the semi-circle example. In this case, a must correspond to thehypotenuse of the right triangle. (Why?) Again we have our choice of how to label thelegs, one side

√a2 − x2 and the other x. With the selection made below, x = a sin θ. In the

circle example we chose to label the legs in the other way because of the geometry involved.Obviously it worked out, but it required us to carry along a minus sign throughout theproblem. The choice below is usually simpler.

....................................................................................................................................................................................................................................................................................................................................................................√

a2 − x2

a

θ

x

x = a sin θ

dx = a cos θ dθ√

a2 − x2 = a cos θ

EXAMPLE 7.3.4. Calculate the following indefinite integral:∫ x2√

25− x2dx.

Use the substitution above with a2 = 25 or a = 5. We use a reduction formula (see theAppendix to these notes) to do the integral.

∫ x2√

25− x2dx =

∫ 25 sin2 θ

5 cos θ· 5 cos θ dθ = 25

∫sin2 θ dθ = 25

∫12 −

12 cos(2θ) dθ

= 252 θ − 25

4 sin(2θ) + c


To solve for θ, look back at the original triangle where x5 = sin θ ⇒ arcsin x

5 = θ and√25−x2

5 = cos θ. The simplest way to finish the problem is to make use of a Double AngleFormula:

sin(2θ) = 2 sin θ cos θ.

Then252 θ − 25

4 sin(2θ) + c = 252 θ − 25

4 (2 sin θ cos θ) + c = 252 arcsin x

5 −252 ·

x5 ·√

25−x2

5 + c

=252

arcsinx5− 1

2· x√

25− x2 + c.

7.4 Additional Examples

In this last section we extend the triangle substitution idea to integrals that at first don’tappear to have anything to do with ‘triangles’ because a square root does not immediatelyappear in them.

EXAMPLE 7.4.1. Calculate the following indefinite integral:∫ 1

(x2 + 6)3/2 dx.

Well there is sort of a square root lurking in the background here. In this case, (x2 + 6)3/2

may be thought of as (√

x2 + 6)3. So the appropriate triangle is:

.................................................................................................................................................................................................................................................................................................................................................................... √

6

√x2 + 6

θ

x

x =√

6 tan θ

dx =√

6 sec2 θ dθ√

x2 + 6 =√

6 sec θ

Notice that one of the sides of the triangle is√

6; we may not always have perfect squares.Using the triangle substitution in the box we have:∫ 1

(x2 + 6)3/2 dx =∫ √

6 sec2 θ

(√

6 sec θ)3dθ =

∫ √6 sec2 θ

6√

6 sec3 θdθ = 1

6

∫ 1sec θ

dθ = 16

∫cos θ dθ

= 16 sin θ + c

=x

6√

x2 + 6+ c.

EXAMPLE 7.4.2. Determine the following indefinite integral:∫ 4

4− x2 dx.

This time we have to think of 4− x2 as (√

4− x2)2 to use a triangle.

....................................................................................................................................................................................................................................................................................................................................................................√

4− x2

2

θ

x

x = 2 sin θ

dx = 2 cos θ dθ√

4− x2 = 2 cos θ

Now we can rewrite the integral.∫ 44− x2 dx =

∫ 4(2 cos θ)2 · 2 cos θ dθ =

∫ 8 cos θ

4 cos2 θdθ = 2

∫sec θ dθ

= 2∫

ln | sec(θ) + tan θ| dθ

= 2 ln | 2√4−x2 + x√

4−x2 |+ c

= 2 ln | 2+x√4−x2 |+ c

= ln | (2+x)2

4−x2 |+ c

= ln | 2+x2−x |+ c

= ln |2 + x| − ln |2− x|+ c.


We will see this same integral again, shortly, and it will be solved very differently.

EXAMPLE 7.4.3. Evaluate the definite integral:∫ 1

1/2

x3√

4x2 − 1dx.

This time the triangle is obvious, but care is required to label the sides correctly.

....................................................................................................................................................................................................................................................................................................................................................................

1

2x

θ

√4x2 − 1

x = 12 sec θ

dx = 12 sec θ tan θ dθ

√4x2 − 1 = tan θ

You could do the indefinite integral and convert back to x to avoid changing the limits.But let’s actually change them

x = 12 = 1

2 sec θ ⇒ 1 = sec θ ⇒ θ = 0x = 1 = 1

2 sec θ ⇒ 2 = sec θ ⇒ cos θ = 12 ⇒ θ = π/3

Now proceed with the substitution (and use the guidelines for integrating powers of sec θ):

∫ 1

1/2

x3√

4x2 − 1dx =

∫ π/3

0

( 12 sec θ)3

tan θ· 1

2 sec θ tan θ dθ =116

∫ π/3

0sec4 θ dθ

=116

∫ π/3

0sec2 θ · sec2 θ dθ

=116

∫ π/3

0(1 + tan2 θ) · sec2 θ dθ

=116

∫ π/3

0sec2 θ +

u2︷︸︸︷tan2 θ

du︷︸︸︷sec2 θ dθ

=116

[tan θ +

tan3 θ

3

] ∣∣∣π/3

0

=116

[√

3 +(√

3)3

3− 0

]=√

38

7.5 Problems

1. Try these problems. A variety of techniques are required, not just triangle substitutions.

(a)∫ √x2 − 9

xdx (b)

∫ 1(1 + x2)3/2 dx (c)

∫ 1√16− x2

dx

(d)∫ √2

0

x2√

4− x2dx (e)

∫ 1

x√

x2 − 9dx (f )

∫x√

1− x2 dx

(g)∫ 4

4 + x2 dx (h)∫ 4x

4 + x2 dx (i)∫ 4

4− x2 dx

(j)∫ √25 + x2

x4 dx (k)∫ 5

0

x√25 + x2

dx (l)∫ 1√

25− x2dx

(m)∫ x3√

25− x2dx (n)

∫ 1√25 + x2

dx

(o)∫ 0

−5

√25− x2 dx (p)

∫ 1

x2√

x2 − 25dx

2. Find the arc length of the parabola f (x) = x2

2 on the interval [0, 1]. You will have to use atrig substitution. Make sure that you switch the limits of the integration.


Answers

1. Caution. . . , it’s easy to have made a typo in these answers. Remember: +c

(a)√

x2 − 9− 3 arctan(√

x2−93

)(b)

x√1 + x2

(c) arcsin(x/4) (d) π2 − 1

(e) 13 arctan

(√x2−93

)(f ) − (1− x2)3/2

3(g) 2 arctan(x/2) (h) 2 ln |4 + x2|

(i) 2 ln∣∣∣∣ 2 + x√

4− x2

∣∣∣∣ (j) − 175

(√25 + x2

x

)3

(k) 5(√

2− 1) (l) arcsin(x/5)

(m) −25√

25− x2 +(25− x2)3/2

3(n) ln

∣∣∣∣∣√

25 + x2

5+

x5

∣∣∣∣∣(o) 25π/4 (p)

√x2 − 2525x

2. Answer:√

2+ln |√

2+1|)2

7.6 Appendix: Common Trigonometric Formulas and Antiderivatives

Below are listed several integral formulas for various powers of trig functions.

1. Degree 2 Sine and Cosine Functions. One simple way to do these is to use trig identi-ties. Know these.

(a)∫

cos2 u du =∫

12 + 1

2 cos 2u du.

(b)∫

sin2 u du =∫

12 −

12 cos 2u du.

2. Low Powers of the Tangent and Secant Functions. These are done with simple identi-ties. Know these.

(a)∫

tan u du =∫ sin u

cos udu = ln | sec u|+ c.

(b)∫

tan2 u du =∫

sec2 u− 1 du = tan u− u + c.

(c)∫

sec u du =∫ sec2 u + sec u tan u

sec u + tan udu = ln | sec u + tan u|+ c.

3. Useful Double Angle Formula: sin(2θ) = 2 sin θ cos θ.

4. Reduction Formulas for Large Powers. These are verified using integration by parts.Repeated application may be necessary.

(a)∫

cosn u du = 1n cosn−1 u sin u + n−1

n∫

cosn−2 u du

(b)∫

sinn u du = − 1n sinn−1 u cos u + n−1

n∫

sinn−2 u du

(c)∫

tann u du = 1n−1 tann−1 u−

∫tann−2 u du

(d)∫

secn u du = 1n−1 secn−2 u tan u + n−2

n−1∫

secn−2 u du

5. Degree 2 Sine and Cosine Functions Again. If we apply the reduction formulas asfor the sine and cosine functions when n = 2, we get a different form of the earlieranswer. These new forms are better to use with indefinite integrals involving trianglesubstitutions because it is easier to convert back from u to the original variable x. (Knoweither these or those in #1.)

(a)∫

cos2 u du = 12 cos u sin u + 1

2∫

1 du = 12 cos u sin u + 1

2 u + c

(b)∫

sin2 u du = − 12 sin u cos u + 1

2∫

1 du = − 12 sin u cos u + 1

2 u + c

Rational Functions and Partial Fractions

Our final integration technique deals with the class of functions known as rationalfunctions. Recall from Calculus I that

DEFINITION 7.1. A rational function1 is a function that is the ratio of two polynomials 1 Here ‘rational’ means ‘ratio’, as in ‘theratio of two polynomials.’

r(x) =p(x)q(x)

,

where p(x) and q(x) are polynomials. (Remember a polynomial has the form

p(x) = anxn + an−1xn−1 + · · ·+ a1x + a0,

where the ai are real constants and n is a non-negative integer and is called the degree of thepolynomial.)

Here are several examples of rational functions; identify the polynomials p(x)and q(x).

• r(x) =x2 + 7

3x5 + 7x

• s(x) =2x3 − 9x + 1

x

• t(x) =1

x2 + 1

• r(x) = 3x−2 + 2x + 1 is also rational because it can be written as a ratio of twopolynomials

r(x) =3x2 + 2x + 1 =

2x3 + x2 + 3x2 .

A couple of non-examples include

• z(x) =x3/2 + 73x + 7x

because the term x3/2 is not a polynomial since the power is not

a non-negative integer.

• s(x) =sin(x2 + 1)

xis not a polynomial since it contains a trigonometric func-

tion.

Remember from Calculus I that rational functions are continuous and differen-tiable at all points in their domains, i.e., at all points where the denominator is not0.

Our main concern in this chapter is to determine when such rational functionscan be integrated. From our earlier work, we know that any interval on which arational function is continuous it is also integrable. But, how do we actually findan antiderivative for the function? For instance, to check that you can integrate

math 131, techniques of integration v partial fractions 2

∫ 11 + x2 dx,

∫ x1 + x2 dx,

∫ 3x2 − 1x3 − x

dx, and∫ 1

1 + xdx. But what about very sim-

ilar looking functions such as∫ 2x + 1

x3 − xdx or

∫ 44− x2 dx? Why are these integrals

not so easy to do? Over the next few sections, we examine a series of special casesof rational functions that we will see are relatively easy to integrate using a tech-nique known as partial fractions.

7.1 Special Cases: Linear Factors with Degree p(x) < q(x)

There are several techniques that can be used integrate rational functions. We willconcentrate on a single technique and a couple of simple variations that work withrational functions of a particular type.

Assume we have a rational function of the form

r(x) =p(x)q(x)

,

where degree of p(x) < degree of q(x) and q(x) can be factored into linear factors(factors of degree 1). Some examples include

•2x + 1x3 − x

=2x + 1

x(x2 − 1)=

2x + 1x(x− 1)(x + 1)

where all the factors in the denominator

are linear and different and degree p(x) = 1 and degree q(x) = 3.

•4

4− x2 =4

(2− x)(2 + x); the two factors in the denominator are linear and

different and degree of p(x) < degree of q(x).

•6x− 2

x2 − 2x− 3=

6x− 2(x− 3)(x + 1)

where all the factors in the denominator are linear

and different and degree of p(x) < degree of q(x).

•x + 4

x3 + x2 =x + 4

x · x(x + 1); the three factors in the denominator are linear and

degree of p(x) < degree of q(x).

On the other hand the rational function

2x + 1x3 + x

=2x + 1

x(x2 + 1)

does not satisfy our criterion above because the denominator contains a factor ofdegree 2, in particular, x2 + 1 cannot be written as a product of linear factors.

We will use a technique called partial fractions to integrate the special rationalfunctions above.2 2 Partial fractions can be used to inte-

grate other types of rational functions.Your text has further examples.

The Key to Partial Fractions

Let’s look at a couple of ordinary fractions and how they can be rewritten in ‘sim-pler’ terms. Notice

120

=14− 1

5or

1021

=13+

17

.


In each case the original fraction has been rewritten in terms of simpler componentfractions. The idea is to do the same for rational functions. For example, can wewrite

44− x2 as

A2− x

+B

2 + x

where 2− x and 2 + x are the linear factors of 4− x2? If we could, then putting thetwo simpler pieces back over a common denominator would give us

44− x2 =

A2− x

+B

2 + x=

2A + Ax + 2B− Bx4− x2 =

(A− B)x + (2A + 2B)4− x2 . (7.1)

The first and last rational functions in (7.1) are equal and have the same denomi-nator. The only way that this can happen is if the numerators are also the same. So(7.1) means

4 = (A− B)x + (2A + 2B). (7.2)

But (7.2) is true if and only if the x-terms on each side are equal and the constantson each side are equal. Since there are no x’s on the left side and A − B on theright, we must have

x’s : 0 = A− B. (7.3)

Comparing the constant terms in the same way we have

constants : 4 = 2A + 2B. (7.4)

From (7.3) we see that A = B and using this in (7.4) gives

4 = 4A.

Thus, A = 1. Putting A = 1 in (7.3) or (7.4) makes B = 1. So we see that

44− x2 =

12− x

+1

2 + x. (7.5)

Check that this is correct! We describe this process by saying that we have rewrit-ten 4

4−x2 as 12−x + 1

2+x using partial fractions.

So what! How do we use this? Well, suppose we need to solve∫ 4

4− x2 dx.

Using (7.5) we can rewrite it and then integrate (using a ‘mental adjustment’):∫ 44− x2 dx =

∫ ( 12− x

+1

2 + x

)dx

= − ln |2− x|+ ln |2 + x|+ c = ln∣∣∣∣2− x2 + x

∣∣∣∣+ c.

7.2 The Easiest Case: Distinct Linear Factors

We can always carry out the same sort of process as above under the followingcircumstances:3 3 This is not so easy to prove in general;

usually one sees the proof of this resultin a graduate-level abstract algebracourse.

The Easiest Case: Let r(x) =r(x)q(x)

be a rational function. Assume that the denominator

q(x) of the rational function factors into distinct linear and the degree of the numer-

ator p(x) is less than the degree of the denominator q(x). Then r(x) =p(x)q(x)

can be

rewritten using partial fractions.


EXAMPLE 7.2.1 (Partial Fractions: Easiest Case). Here’s another example that uses partial frac-tions. Determine ∫ 3

x2 + 3x + 2dx.

SOLUTION. This is not an integral that we can immediately do, even with integrationby parts. 4 So we try partial fractions. Notice that the degree of the numerator is less 4 Don’t confuse the terms partial fractions

with integration by parts.than the degree of the denominator (0 < 2) and the denominator factors into distinctlinear factors: (x + 1)(x + 2). We form the partial fractions A

x+1 and Bx+2 , where A and

B are constants. We solve for A and B as we did in the previous example.

3x2 + 3x + 2

=3

(x + 1)(x + 2)=

Ax + 1

+B

x + 2.

By putting the last terms together again, we get

3x2 + 3x + 2

=Ax + 2A + Bx + B

x2 + 3x + 2=

(A + B)x + (2A + B)x2 + 3x + 2

.

Since the denominators are the same, the numerators must be the same, too. In par-ticular there are as many x’s on the left side as on the right. There are none on the leftand A + B on the right side. Similarly for the constants.

x’s: 0 = A + Bconstants: 3 = 2A + B

Subtracting the first equation from the second gives

3 = A.

Substituting A = 3 into first or second equation makes B = −3. So we see that

3x2 + 3x + 2

=3

x + 1− 3

x + 2.

Now on to integration:∫ 3x2 + 3x + 2

dx =∫ ( 3

x + 1− 3

x + 2

)dx

= 3 ln |x + 1| − 3 ln |x + 2|+ c = 3 ln∣∣∣∣ x + 1

x + 2

∣∣∣∣+ c.

EXAMPLE 7.2.2 (Partial Fractions: Easiest Case). Determine∫ 6x− 2x2 − 2x− 3

dx.

SOLUTION. This is not an integral that we can immediately do with substitution orintegration by parts. So we try partial fractions. The degree of the numerator is lessthan the degree of the denominator (1 < 2) and the denominator factors into distinctlinear factors: (x− 3)(x + 1). We form the partial fractions A

x−3 and Bx+1 , where A and

B are constants.

6x− 2x2 − 2x− 3

=6x− 2

(x− 3)(x + 1)=

Ax− 3

+B

x + 1=

Ax + A + Bx− 3Bx2 − 2x− 3

.

So comparing the numerators of the first and last functions

x’s: 6 = A + Bconstants: −2 = A− 3B

Subtracting the second equation from the first gives

8 = 4B.

So B = 2. Using this in the first or second equation makes A = 4. Now on to integra-tion (using an adjustment):∫ 6x− 2

x2 − 2x− 3dx =

∫ ( 4x− 3

+2

x + 1

)dx = 4 ln |x− 3|+ 2 ln |x + 1|+ c.


YOU TRY IT 7.1. Does it matter which letters you put over each linear factor? What wouldthe numerator in the original integral have to be to make the problem a substitution prob-lem? Answers to you try it 7.1 : No. It

would have to be a multiple of 2x− 2.EXAMPLE 7.2.3 (Partial Fractions: Easiest Case). Here’s another quick example. Determine∫ 2x + 5

x2 + 2x− 8dx.

SOLUTION. Notice this is not quite a substitution integral, but partial fractions willwork. The degree of the numerator is less than the degree of the denominator (1 < 2)and the denominator factors into distinct linear factors: (x− 2)(x + 4).

2x + 5x2 + 2x− 8

=2x + 5

(x− 2)(x + 4)=

Ax− 2

+B

x + 4=

Ax + 4A + Bx− 2Bx2 + 2x− 8

.

Comparing the numerators of the first and last functions and solving for A and Bgives

x’s: 2 = A + B ⇒ 4 = 2A + 2Bconstants: 5 = 4A− 2B 5 = 4A− 2B

9 = 6A

So A = 32 and B = 1

2 . The integration becomes:

∫ 2x + 5x2 + 2x− 8

dx =∫ ( 3

2x− 2

+12

x + 4

)dx = 3

2 ln |x− 2|+ 12 ln |x + 4|+ c.

YOU TRY IT 7.2. What would the numerator in the original integral have to be to make theproblem a substitution problem?

7.3 A Complication: Higher Degrees

The first complication that arises is is that the denominator of the rational functionmay factor into three or more distinct linear factors. The solution method worksthe same way as above, but it may be more complicated to find the constants.

EXAMPLE 7.3.1 (Partial Fractions: Three Distinct Linear Factors). Determine∫ 2x2 − 6x + 2x3 − 3x2 + 2x

dx.

SOLUTION. Check that this is not quite a substitution integral. However, the degreeof the numerator is less than the degree of the denominator (2 < 3) and the denom-inator factors into distinct three linear factors: x(x2 − 3x + 2) = x(x − 1)(x − 2). Weform the partial fractions with a different constant for each linear factor:

2x2 − 6x + 2x3 − 3x2 + 2x

=2x2 − 6x + 2

x(x− 1)(x− 2)=

Ax+

Bx− 1

+C

x− 2

=A(x− 1)(x− 2) + Bx(x− 2) + Cx(x− 1)

x3 − 3x2 + 2x

=Ax2 − 3Ax + 2A + Bx2 − 2B + Cx2 − Cx

x3 − 3x2 + 2x.

Compare like terms in the numerators of the first and last functions.

x2’s: 2 = A + B + C ⇒ 1 = B + Cx’s: −6 = −3A− 2B− C ⇒ −3 = −2B− C

const: 2 = 2A ⇒ A = 1 −2 = −B

Notice that we used the value A = 1 to simplify the first two equations. It follows thatB = 2 and C = −1. The integration becomes:∫ 2x2 − 6x + 2

x3 − 3x2 + 2xdx =

∫ 1x+

2x− 1

− 1x− 2

dx = ln |x|+ 2 ln |x− 1| − ln |x− 2|+ c.



EXAMPLE 7.3.2 (Partial Fractions: Three Distinct Linear Factors). Determine∫ x2 + 4x− 1x3 − x

dx.

SOLUTION. Check that this is not a substitution integral. However, the degree of thenumerator is less than the degree of the denominator (2 < 3) and the denominatorfactors into distinct three linear factors: x(x2 − 1) = x(x − 1)(x + 1). We form thepartial fractions with a different constant for each linear factor:

x2 + 4x− 1x3 − x

=Ax+

Bx− 1

+C

x + 1=

A(x2 − 1) + Bx(x + 1) + Cx(x− 1)x3 − x

=Ax2 − A + Bx2 + B + Cx2 − Cx

x3 − x.


x2’s: 1 = A + B + C ⇒ 0 = B + Cx’s: 4 = B− C ⇒ 4 = B− C

const: −1 = −A ⇒ A = 1 4 = 2B

Notice that we used the value A = 1 to simplify the first two equations. It follows thatB = 2 and C = −2. The integration becomes:∫ x2 + 4x− 1

x3 − xdx =

∫ 1x+

2x− 1

− 2x + 1

dx = ln |x|+ 2 ln |x− 1| − 2 ln |x + 1|+ c

= ln |x|+ 2 ln∣∣∣∣ x− 1

x + 1

∣∣∣∣+ c.

EXAMPLE 7.3.3 (Partial Fractions: Three Distinct Linear Factors). Determine∫ 4x + 28(x + 1)(x2 − 4x + 3)

dx.

SOLUTION. Check that this is not a substitution integral. However, the degree of thenumerator is less than the degree of the denominator (1 < 3) and the denominatorfactors into distinct three linear factors: (x + 1)(x2 − 4x + 3) = (x + 1)(x− 1)(x− 3).We form the partial fractions with a different constant for each linear factor:

4x + 28(x + 1)(x2 − 4x + 3)

=A

x + 1+

Bx− 1

+C

x− 3=

A(x2 − 4x + 3) + B(x2 − 2x− 3) + C(x2 − 1)(x + 1)(x2 − 4x + 3)

=(A + B + C)x2 + (−4A− 2B)x + 3A− 3B− C

(x + 1)(x2 − 4x + 3).


x2’s: 0 = A + B + C ⇒ C = 5x’s: 4 = −4A− 2B ⇒ A = 3

const: 28 = 3A− 3B− CAdd all: 32 = − 4B ⇒ B = −8

The integration becomes:∫ 4x + 28(x + 1)(x2 − 4x + 3)

dx =∫ 3

x + 1− 8

x− 1+

5x− 3

dx

= 3 ln |x− 1| − 8 ln |x + 1|+ 5 ln |x− 3|+ c



7.4 A Second Complication: Repeated Factors

Another complication that can arise is that the denominator of the rational func-tion factors into linear factors, but the factors are not distinct. That is, there arerepeated linear factors. The solution in this situation is to include a term for everypower of every linear factor that divides the denominator.

EXAMPLE 7.4.1 (Partial Fractions: Repeated Linear Factors). Determine∫ 3x2 − 7x + 2x3 − 2x2 + x

dx.

SOLUTION. This is not a substitution integral. However, the degree of the numeratoris less than the degree of the denominator (2 < 3) and the denominator factors intorepeated linear factors:

x(x2 − 2x + 1) = x(x− 1)(x− 1) = x(x− 1)2.

We form the partial fractions with a different constant for each power of each factorthat divides the denominator. Note: There are terms for both x − 1 and (x − 1)2.Warning! Look very carefully at the numerator in the step where the common denom-inator is formed.

3x2 − 7x + 2x(x− 1)2 =

Ax+

Bx− 1

+C

(x− 1)2 =A(x− 1)2 + Bx(x− 1) + Cx

x(x− 1)2

=Ax2 − 2Ax + A + Bx2 − Bx + Cx

x(x− 1)2 .


x2’s: 3 = A + B ⇒ 1 = Bx’s: −7 = −2A− B + C ⇒ C = −2const: 2 = A ⇒ A = 2

Notice that we used the value A = 2 to simplify the first two equations. The integra-tion becomes:∫ 3x2 − 7x + 2

x3 − 2x2 + xdx =

∫ 2x+

1x− 1

− 2(x− 1)2 dx

= 2 ln |x|+ ln |x− 1|+ 2(x− 1)−1 + c.

Be careful! The final integral requires the power rule and a mini-substitution.


EXAMPLE 7.4.2 (Partial Fractions: Repeated Linear Factors). Determine∫ 3x2 − 2x− 3x3 − x2 dx.

SOLUTION. This is not quite a substitution integral. However, the degree of the nu-merator is less than the degree of the denominator (2 < 3) and the denominatorfactors into repeated linear factors: x2(x − 1). We form the partial fractions with adifferent constant for each power of each factor that divides the denominator—thereare terms for both x and x2. Look very carefully at the numerator in the step wherethe common denominator is formed.

3x2 − 2x− 3x2(x− 1)

=Ax+

Bx2 +

Cx− 1

=Ax(x− 1) + B(x− 1) + Cx2

x2(x− 1)

=Ax2 − Ax + Bx− B + Cx2

x2(x− 1).



x2’s: 3 = A + C ⇒ C = −2x’s: −2 = −A + B ⇒ A = 5const: −3 = −B ⇒ B = 3

Notice that we used the value B = 3 to simplify the first two equations. The integra-tion becomes:∫ 3x2 − 2x− 3

x3 − x2 dx =∫ 5

x+

3x2 −

2x− 1

dx = 5 ln |x| − 3x−1 − 2 ln |x− 1|+ c.


YOU TRY IT 7.7. Determine Answer to you try it 7.7 : 2 ln |x + 1| −2 ln |x− 1| − 3

x−1 .∫ 7− x

(x + 1)(x− 1)2 dx.

EXAMPLE 7.4.3 (Partial Fractions: Repeated Linear Factors). Here’s a different one: Determine∫ x2

(x + 1)3 dx.

SOLUTION. The degree of the numerator is less than the degree of the denominator(2 < 3) and the denominator factors into repeated linear factors: (x + 1)3 = (x +

1)(x + 1)(x + 1). We form the partial fractions with a different constant for eachpower of each factor that divides the denominator: (x + 1), (x + 1)2, and (x + 1)3.Look very carefully at the numerator in the step where the common denominator isformed.

x2

(x + 1)3 =A

x + 1+

B(x + 1)2 +

C(x + 1)3 =

A(x + 1)2 + B(x + 1) + C(x + 1)3

=Ax2 + 2Ax + A + Bx + B + C

(x + 1)3 .


x2’s: 1 = A ⇒ A = 1x’s: 0 = 2A + B ⇒ B = −2const: 0 = A + B + C ⇒ C = 1

Notice that we used the value A = 1 to simplify the last two equations. The integra-tion becomes:∫ x2

(x + 1)3 dx =∫ 1

x + 1− 2

(x + 1)2 +1

(x + 1)3 dx

= ln |x + 1|+ 2(x + 1)−1 − 12 (x + 1)−2 + c.

Alternative Solution. We could have used a u-substitution to solve this problem. Letu = x + 1. Then du = dx. Since u = x + 1, then x = u − 1 so x2 = (u − 1)2 =

u2 − 2u + 1. Therefore∫ x2

(x + 1)3 dx =∫ u2 − 2u + 1

u3 du =∫ 1

u− 2

u2 +1

u3 du

= ln |u| − 2u−1 − 12 u−2 + c

= ln |x + 1|+ 2(x + 1)−1 − 12 (x + 1)−2 + c,

just as above. Which method would you have used? Why?

7.5 Problems


1. Try integrating these rational functions (answers below). These are a bit harder than

those in the text. Some have three factors. Others have repeated factors.

(a)∫ 4t2 − 3t− 4

t3 − t2 − 2tdt (b)

∫ t + 7(t + 1)(t2 − 4t + 3)

dt (c)∫ x + 6

x2 − x− 6dx

(d)∫ x

(x− 1)(x + 1)(x + 2)dx (e)

∫ 2x2

(x− 1)2(x + 1)dx (f )

∫ −4x + 4(x− 2)2x

dx

(g)∫ 2x

(x− 2)(x2 − 1)dx (h)

∫ 2x− 1(x + 2)2 dx

2. Try these similar looking problems.

(a)∫ 10

25 + x2 dx (b)∫ 10x

25 + x2 dx (c)∫ 10

25− x2 dx

3. Have you finished all the ones above? Do these similar looking integrals.

(a)∫ 4√

4 + x2dx (b)

∫ 4√4− x2

dx

(c)∫ 4x

(4 + x2)3/2 dx (d)∫ 2

−2

√4− x2 dx

Answers to Practice Problems

1. (a)∫ 2

t +1

t−2 + 1t+1 dt = 2 ln |t|+ ln |t− 2|+ ln |t + 1|+ c

(b)∫ 3/4

t+1 + 5/4t−3 −

2t−1 dt = 3

4 ln |t + 1|+ 54 ln |t− 3| − 2 ln |t− 1|+ c

(c)∫ 9/5

x−3 −4/5x+2 dt = x + 9

5 ln |x− 3| − 45 ln |x + 2|+ c

(d)∫ 1/6

x−1 + 1/2x+1 −

2/3x+2 dx = 1

6 ln |x− 1|+ 12 ln |x + 1| − 2

3 ln |x + 2|+ c

(e)∫ 3/2

x−1 + 1(x−1)2 +

1/2x+1 dx = 3

2 ln |x− 1| − (x− 1)−1 + 12 ln |x + 1|+ c

(f )∫− 1

x−2 −2

(x−2)2 +1x dx = − ln |x− 2|+ 2(x− 2)−1 + ln |x|+ c

(g)∫ 2

x+2 −5

(x+2)2 dx = 2 ln |x + 2|+ 5x+2 + c.

2. All “+c".

(a) 2 arctan(x/5) (b) 5 ln |25 + x2| (c) ln∣∣∣∣ x + 5x− 5

∣∣∣∣

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