Some Standard Integration Techniques

S. F. Ellermeyer

January 11, 2005

1 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) tells us that if a function, f ,is continuous on the interval [a, b] and the function F is any antiderivativeof f on [a, b], then Z b

a

f (x) dx = F (b)− F (a) .In order to apply this result, we need to be able to find an antiderivative, F ,of the integrand, f . Sometimes this is impossible, but there are many casesin which it can be done.We will employ the idea of the “indefinite integral” of f , written asZ

f (x) dx,

to mean the set of all antiderivatives of f on some interval, I. If F is anyparticular antiderivative of f on the interval I, thenZ

f (x) dx = F (x) + C (where C can be any real constant).

Since the interval, I, does not appear anywhere in the notation for the indefi-nite integral, it is not a very good notation. This usually causes no problems,however, in a course such as this where we mostly consider functions whoseantiderivatives do not depend on which interval in the domain of f we areconsidering. For example, there is no ambiguity when we writeZ

x2 dx =1

3x3 + C

1

or Zcos (x) dx = sin (x) + C

because the corresponding differentiation formulas

d

dx

µ1

3x3 + C

¶= x2

andd

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

are correct on any subinterval of (−∞,∞). One notable exception (thatwe do encounter very often) is the function f (x) = 1/x. In particular, thestatement Z

1

xdx = ln (x) + C

is correct if we are considering the function f (x) = 1/x with domain asubinterval of (0,∞), but it is not correct if the domain under considerationis a subinterval (−∞, 0). In the latter case, a correct statement isZ

1

xdx = ln (−x) + C.

We can always be safe (that is, avoid any possible confusion) by using theformula Z

1

xdx = ln |x|+ C.

This last statement is correct in both the case that the domain of f (x) = 1/xis a subinterval (0,∞) and the case that the domain of f (x) = 1/x is asubinterval of (−∞, 0).

2 A Basic Beginning Toolbox of Integrals

Below is a list of some basic integrals. These are integrals that should bememorized. All of the integration techniques that we use to compute morecomplicated integrals are aimed at reducing the more complicated integralsto one of the forms in the basic list.

2

1. If n is any fixed real number (except −1), thenZxn dx =

1

n+ 1xn+1 + C.

2. Zx−1 dx = ln |x|+ C

3. Zex dx = ex + C

4. Zcos (x) dx = sin (x) + C

5. Zsin (x) dx = − cos (x) + C

6. Zsec2 (x) dx = tan (x) + C

7. Zcsc2 (x) dx = − cot (x) + C

8. Zsec (x) tan (x) dx = sec (x) + C

9. Zcsc (x) cot (x) dx = − csc (x) + C

10. Z1√1− x2 dx = arcsin (x) + C

11. Z1

1 + x2dx = arctan (x) + C.

3

3 Integration by Substitution

The Chain Rule tells us that is F and g are differentiable functions, then

d

dx(F (g (x))) = F 0 (g (x)) · g0 (x) .

Thus, ZF 0 (g (x)) · g0 (x) dx = F (g (x)) + C.

If we make the substitutionsu = g (x)

anddu = g0 (x) dx,

then we can write the above integration formula asZF 0 (u) du = F (u) + C

This leads to the idea of “integration by substitution”.

Example 1 Evaluate the indefinite integralZx2¡4x3 + 5

¢8dx.

Solution 2 Make the substitution

u = 4x3 + 5

du = 12x2dx

and note thatx2dx =

1

12du.

The above integral can then be written asZ1

12u8 du.

Since Z1

12u8 du =

1

12· 19u9 + C,

we see that Zx2¡4x3 + 5

¢8dx =

1

108

¡4x3 + 5

¢9+ C.

4

Example 3 Evaluate the definite integralZ π/4

0

tan2 (x) sec2 (x) dx.

Solution 4 Making the substitution

u = tan (x)

du = sec2 (x) dx,

we see thatZtan2 (x) sec2 (x) dx =

Zu2 du =

1

3u3 + C =

1

3tan3 (x) + C.

By the FTC, we then haveZ π/4

0

tan2 (x) sec2 (x) dx =1

3tan3

³π4

´− 13tan3 (0) =

1

3.

(Recall that tan¡π4

¢= 1 and tan (0) = 0.)

Exercise 5 Evaluate the following indefinite or definite integrals.

1. Zsin¡3x2 + 2x− 1¢ · (3x+ 1) dx

2. Z ln(3)

0

ex√ex + 4 dx

3. Z3x2 + 8x− 6

x3 + 4x2 − 6x− 2 dx

4. Z π/4

0

tan (x) dx

5. Zex

e2x + 1dx

Hint: e2x = (ex)2.

5

4 Integration By Parts

The Product Rule tells us that if f and g are differentiable functions, then

d

dx(f (x) · g (x)) = f (x) · g0 (x) + f 0 (x) · g (x) .

Thus Z(f (x) · g0 (x) + f 0 (x) · g (x)) dx = f (x) · g (x) + C

or Zf (x) · g0 (x) dx+

Zf 0 (x) · g (x) dx = f (x) · g (x) + C

Making the substitutions

u = f (x)

du = f 0 (x) dxv = g (x)

dv = g0 (x) dx,

we can write the above integration formula asZudv +

Zv du = uv + C

or as Zudv = uv −

Zv du+ C.

(It is not really necessary to write the “+C” at the end of the above formula,since the indefinite integral on the right automatically includes a “+C”.)The above formula is what we call the formula for integration by parts. It

is used in cases where the integralRv du is easy to compute, but the integralR

udv is not. Specifically, it is used when we want to computeRudv, and

we do this via computingRv du.

Example 6 Evaluate the indefinite integralZx cos (x) dx.

6

Solution 7 The above integral has the formRudv where

u = x dv = cos (x) dxdu = dx v = sin (x)

.

Using the integration by parts formula, we obtainZx cos (x) dx =

Zudv

= uv −Zv du

= x sin (x)−Zsin (x) dx

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

We conclude that Zx cos (x) dx = x sin (x) + cos (x) + C.

Sometimes, it is necessary to use integration by parts more than once, asin the next example.

Example 8 Evaluate the indefinite integralZex cos (x) dx.

Solution 9 Make the following substitutions

u = ex dv = cos (x) dxdu = exdx v = sin (x)

.

Then Zex cos (x) dx =

Zudv

= uv −Zv du

= ex sin (x)−Zex sin (x) dx.

7

We now see thatZex cos (x) dx = ex sin (x)−

Zex sin (x) dx. (1)

At this point, it appears that integration by parts has perhaps been of nohelp, because

Rv du appears to be no easier than the original problem,

Ru dv.

However, we continue (using integration by parts again) with the new problemZex sin (x) dx.

Here we make the substitutions

u = ex dv = sin (x) dxdu = exdx v = − cos (x)

and obtain Zex sin (x) dx = uv −

Zv du

= −ex cos (x) +Zex cos (x) dx.

Thus, Zex sin (x) dx = −ex cos (x) +

Zex cos (x) dx. (2)

Now, combining the above results (equations (1) and (2)), we observe thatZex cos (x) dx = ex sin (x)−

Zex sin (x) dx

= ex sin (x)−µ−ex cos (x) +

Zex cos (x) dx

¶= ex sin (x) + ex cos (x)−

Zex cos (x) dx

or, in summary,Zex cos (x) dx = ex sin (x) + ex cos (x)−

Zex cos (x) dx. (3)

8

AddingRex cos (x) dx to both sides of equation (3) gives us

2

Zex cos (x) dx = ex sin (x) + ex cos (x) .

We then divide both sides of the above equation by 2 to obtainZex cos (x) dx =

1

2ex (sin (x) + cos (x)) + C.

Exercise 10 Evaluate the following integrals.

1. Zx ln (x) dx

Hint: Let u = ln (x) and dv = x dx.

2. Zx sin (x) dx

3. Zex sin (x) dx

4. Z 1

0

xe2x dx

5. Zx sin (4x+ 2) dx

6. Z π/2

0

x2 cos (3x) dx

7. Zln (x) dx

9

8. Z(ln (x))2 dx

Hint: Use your result from exercise 7.

9. Zarctan (x) dx

10. Zx arctan (x) dx

Hint: Use your result from exercise 9 and you may also find it usefulto use the fact that

x2

1 + x2= 1− 1

1 + x2.

11. Ze3x cos (4x) dx

5 Trigonometric Integrals

Let’s consider how to evaluate integrals of the formZsinm (x) cosn (x) dx

where m and n are positive integers. The right approach to take depends onwhether m and n are even or odd integers.

Example 11 Evaluate Zsin5 (x) cos4 (x) dx.

Solution 12 The key to evaluating this integral is to notice that we have anodd power of sine in the integrand. We would like to “reserve” a factor of

10

sin (x) so that we can make the substitution u = cos (x), du = − sin (x) dx.We can do this by rewriting the integrand as

sin5 (x) cos4 (x) = cos4 (x) sin4 (x) sin (x)

= cos4 (x)¡sin2 (x)

¢2sin (x)

= cos4 (x)¡1− cos2 (x)¢2 sin (x)

= cos4 (x)¡1− 2 cos2 (x) + cos4 (x)¢ sin (x)

=¡cos4 (x)− 2 cos6 (x) + cos8 (x)¢ sin (x) .

Now using the substitution u = cos (x), du = − sin (x) dx, we haveZsin5 (x) cos4 (x) dx =

Z ¡cos4 (x)− 2 cos6 (x) + cos8 (x)¢ sin (x) dx

= −Z ¡

u4 − 2u6 + u8¢ du=

Z ¡−u4 + 2u6 − u8¢ du= −1

5u5 +

2

7u7 − 1

9u9 + C

= −15cos5 (x) +

2

7cos7 (x)− 1

9cos9 (x) + C.

The general rule to be learned from the above example is that if eitherm or n is an odd integer in the integralZ

sinm (x) cosn (x) dx,

then “reserve” one factor of sine or cosine (whichever has the odd power)and use the Pythagorean identity sin2 (x) + cos2 (x) = 1 to rewrite the restof the integrand in terms of either sine or cosine (whichever has not been“reserved”). If both m and n are odd, then we can choose to reserve eithera factor of sine or a factor of cosine.If both m and n are even, then we must make use of the trigonometric

identities

sin2 (x) =1− cos (2x)

2

cos2 (x) =1 + cos (2x)

2.

11

The identitysin (2x) = 2 sin (x) cos (x)

might also come in handy.This idea is illustrated in the following example.

Example 13 Evaluate Zsin4 (x) cos2 (x) dx.

Solution 14 We rewrite the integrand as follows:

sin4 (x) cos2 (x) = (sin (x) cos (x))2 sin2 (x)

=

µ1

2sin (2x)

¶2µ1− cos (2x)

2

¶=

1

4sin2 (2x)

µ1

2− 12cos (2x)

¶=

1

8sin2 (2x)− 1

8sin2 (2x) cos (2x) .

At this point, note that the integral of

1

8sin2 (2x) cos (2x)

will not be hard to do because it has an odd power of cosine. We will leavethis term as it is, but we still need to put the term

1

8sin2 (2x)

into a form that we can integrate. Continuing, we have

sin4 (x) cos2 (x) =1

8sin2 (2x)− 1

8sin2 (2x) cos (2x)

=1

8

µ1− cos (4x)

2

¶− 18sin2 (2x) cos (2x)

=1

8

µ1

2− 12cos (4x)

¶− 18sin2 (2x) cos (2x)

=1

16− 1

16cos (4x)− 1

8sin2 (2x) cos (2x) .

12

We can now compute the desired integral by breaking it into three integrals:Zsin4 (x) cos2 (x) dx =

Z µ1

16− 1

16cos (4x)− 1

8sin2 (2x) cos (2x)

¶dx

=

Z1

16dx− 1

16

Zcos (4x) dx

−18

Zsin2 (2x) cos (2x) dx

=1

16x− 1

64sin (4x)− 1

8

Zsin2 (2x) cos (2x) dx.

To do the last integral (which contains an odd power of cosine), we use thesubstitution u = sin (2x), du = 2 cos (2x) dx. This gives usZ

sin2 (2x) cos (2x) dx =1

2

Zu2 du

=1

6u3 + C

=1

6sin3 (2x) + C.

Our final result isZsin4 (x) cos2 (x) dx =

1

16x− 1

64sin (4x)− 1

48sin3 (2x) + C.

We will now check our result graphically: Below are graphs (drawn usingMaple) of the functions

F (x) =1

16x− 1

64sin (4x)− 1

48sin3 (2x)

andf (x) = sin4 (x) cos2 (x) .

By comparing these graphs, it looks reasonable that F 0 (x) = f (x).

13

Exercise 15 Evaluate the following integrals:

1. Zcos4 (x) sin (x) dx

2. Zcos4 (x) dx

3. Zsin3 (x) cos3 (x) dx

4. Zsin5 (x) dx

5. Zsin2 (x) cos2 (x) dx

14

6 Trigonometric Substitution

In dealing with integrals whose integrands contain terms of the form√a2 − x2,√

a2 + x2, or√x2 − a2 (where a is a constant), it is often helpful to make a

trigonometric substitution.If we have the term

√a2 − x2, then the substitution to try is

x = a sin (θ) .

In this case we havedx = a cos (θ) dθ

and alsox2 = a2 sin2 (θ)

so√a2 − x2 =

qa2 − a2 sin2 (θ)

=qa2¡1− sin2 (θ)¢

= apcos2 (θ)

= a cos (θ) .

(This removes the square root from the problem.) This is illustrated in thefollowing example.

Example 16 Use integration to show that the area of a circle of radius r isπr2.

Solution 17 The equation for the circle of radius r centered at the origin isx2 + y2 = r2. The equation for the upper half of this circle is y =

√r2 − x2.

Thus, the area of the circle is

A = 2

Z r

−r

√r2 − x2 dx.

To evaluate this integral, we make the trigonometric substitution

x = r sin (θ) .

This gives usdx = r cos (θ) dθ

15

and also gives us

√r2 − x2 =

qr2 − r2 sin2 (θ) = r cos (θ) .

Now we have

A = 2

Z r

−r

√r2 − x2 dx

= 2

Z π/2

−π/2r cos (θ) r cos (θ) dθ

= 2r2Z π/2

−π/2cos2 (θ) dθ

= 2r2Z π/2

−π/2

1 + cos (2θ)

2dθ

= 2r2Z π/2

−π/2

µ1

2+1

2cos (2θ)

¶dθ

= 2r2µ1

2θ +

1

4sin (2θ)

¶¯̄̄̄θ=π/2θ=−π/2

= 2r2³³π4+ 0´−³−π4+ 0´´

= 2r2³π2

´= πr2.

The relevant trigonometric substitution for dealing with integrands thatcontain terms of the form

√a2 + x2 is x = a tan (θ) and the relevant one for

dealing with terms of the form√x2 − a2 is x = a sec (θ).

Example 18 Evaluate the integralZ1√4 + x2

dx.

Solution 19 We make the trigonometric substitution

x = 2 tan (θ) .

16

This gives usdx = 2 sec2 (θ) dθ

and

√4 + x2 =

q4 + 4 tan2 (θ)

=q4 (1 + tan2 (θ))

= 2psec2 (θ)

= 2 sec (θ) .

We now obtainZ1√4 + x2

dx =

Z1

2 sec (θ)· 2 sec2 (θ) dθ =

Zsec (θ) dθ.

The integral Zsec (θ) dθ

is very tricky to do (if you don’t know the trick) and perhaps this trick is onethat you should remember! Here’s how to do it: We observe that

sec (θ) =sec (θ)

1· sec (θ) + tan (θ)sec (θ) + tan (θ)

=sec2 (θ) + sec (θ) tan (θ)

sec (θ) + tan (θ).

Thus Zsec (θ) dθ =

Zsec2 (θ) + sec (θ) tan (θ)

sec (θ) + tan (θ)dθ.

We now make the simple substitution

u = sec (θ) + tan (θ) .

Thendu =

¡sec2 (θ) + sec (θ) tan (θ)

¢dθ

and we haveZsec2 (θ) + sec (θ) tan (θ)

sec (θ) + tan (θ)dθ =

Z1

udu = ln |u|+ C.

17

To summarize from the beginning, we haveZ1√4 + x2

dx =

Zsec (θ) dθ =

Z1

udu = ln |u|+ C.

Our answer is in terms of u, but we would like it to be in terms of x, so weneed to substitute back:Z

1√4 + x2

dx = ln |u|+ C= ln |sec (θ) + tan (θ)|+ C= ln

¯̄̄̄q1 + tan2 (θ) + tan (θ)

¯̄̄̄+ C

= ln

¯̄̄̄¯̄s1 +

µ1

2x

¶2+1

2x

¯̄̄̄¯̄+ C.

In conclusion, Z1√4 + x2

dx = ln

¯̄̄̄¯r1 +

1

4x2 +

1

2x

¯̄̄̄¯+ C.

We can make this answer look nicer: Note thatr1 +

1

4x2 =

r4 + x2

4=1

2

√4 + x2

so r1 +

1

4x2 +

1

2x =

1

2

√4 + x2 +

1

2x =

1

2

³√4 + x2 + x

´and

ln

¯̄̄̄¯r1 +

1

4x2 +

1

2x

¯̄̄̄¯ = ln

¯̄̄̄1

2

³√4 + x2 + x

´¯̄̄̄= ln

¯̄̄̄1

2

¯̄̄̄+ ln

¯̄̄√4 + x2 + x

¯̄̄.

The ln |1/2| can be “absorbed” in the integration constant C. Also, since√4 + x2 + x is always positive, we can get rid of the absolute value. This

gives us the resultZ1√4 + x2

dx = ln³√4 + x2 + x

´+ C.

18

Let us check that this result is correct by differentiation:

d

dx

³ln³√4 + x2 + x

´´=

1√4 + x2 + x

·µ1

2

¡4 + x2

¢−1/2 · 2x+ 1¶=

1√4 + x2 + x

·µ

x√4 + x2

+ 1

¶=

1√4 + x2 + x

·Ãx+√4 + x2√

4 + x2

!=

1√4 + x2

.

Exercise 20 Evaluate the following integrals:

1. Z1√9− x2 dx

2. Z1√9 + x2

dx

3. Z √25− x2 dx

4. Z √4 + x2 dx

5. Z1√x2 − 9 dx

7 Partial Fraction Decomposition

To evaluate integrals of the formZp(x)

q(x)dx

19

where p and q are polynomials with the degree of p less than the degree ofq, we sometimes first need to write p (x) /q (x) as a sum of rational functionswhose denominators are powers of linear and irreducible quadratic factors ofq. This is called doing a “partial fraction decomposition” of the integrand.We illustrate this technique by looking at some examples.

Example 21 Consider the problem of computing the indefinite integralZx+ 1

x3 + 2x2 − 3x dx. (4)

In factored form the integrand is

x+ 1

x(x− 1)(x+ 3) .

We attempt to decompose this rational function as

x+ 1

x(x− 1)(x+ 3) =A

x+

B

x− 1 +C

x+ 3(5)

where A, B, and C are constants to be determined. To determine the appro-priate constants, we multiply both sides of equation (5) by the denominatorof the left hand side to obtain

x+ 1 = A(x− 1)(x+ 3) +Bx(x+ 3) + Cx(x− 1). (6)

Substitution of x = 0 into (6) gives

0 + 1 = A(0− 1)(0 + 3) +B(0)(0 + 3) + C(0)(0− 1)from which we conclude that A = −1/3.Substitution of x = 1 into (6) gives

1 + 1 = A(1− 1)(1 + 3) +B(1)(1 + 3) + C(1)(1− 1)which yields that B = 1/2.Similarly, substitution of x = −3 into (6) gives C = −1/6.Substituting the values of A, B, and C which have been found back into

(5), we obtain the partial fraction decomposition

x+ 1

x(x− 1)(x+ 3) =−1/3x

+1/2

x− 1 +−1/6x+ 3

, (7)

20

which can also be written as

x+ 1

x(x− 1)(x+ 3) = −1

6

µ2

x− 3

x− 1 +1

x+ 3

¶.

The indefinite integral (4) can now be computed:Zx+ 1

x3 + 2x2 − 3x dx = −16

Z µ2

x− 3

x− 1 +1

x+ 3

¶dx

= −16(2 ln |x|− 3 ln |x− 1|+ ln |x+ 3|) + C

= −16

¡lnx2 − ln |x− 1|3 + ln |x+ 3|¢+ C

= −16ln

µx2|x+ 3||x− 1|3

¶+ C

Example 22 Let us write the rational expression

1

x(x+ 1)(x+ 2)3

in partial fraction form.Like the rational expression which was the subject of Example 21, the

rational expression1

x(x+ 1)(x+ 2)3

has a denominator which is a product of linear factors. Since the factor x+2has multiplicity three, the partial fraction decomposition will have the form

1

x(x+ 1)(x+ 2)3=A

x+

B

x+ 1+

C

x+ 2+

D

(x+ 2)2+

E

(x+ 2)3. (8)

Multiplication of both sides of (8) by the denominator of the left hand sideyields

1 = A(x+ 1)(x+ 2)3 +Bx(x+ 2)3 + Cx(x+ 1)(x+ 2)2 (9)

+Dx(x+ 1)(x+ 2) +Ex(x+ 1).

When x = 0 is substituted into (9), we obtain

1 = A(0 + 1)(0 + 2)3,

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which implies that A = 1/8.Substitution of x = −1 and x = −2 yield, respectively, that B = −1 and

E = 1/2Hence, equation (9) becomes

1 =1

8(x+ 1)(x+ 2)3 − x(x+ 2)3 + Cx(x+ 1)(x+ 2)2 (10)

+Dx(x+ 1)(x+ 2) +1

2x(x+ 1).

When x = −3 is substituted into (10), we obtain

1 =1

8(−3 + 1)(−3 + 2)3 − (−3)(−3 + 2)3 + C(−3)(−3 + 1)(−3 + 2)2

+D(−3)(−3 + 1)(−3 + 2) + 12(−3)(−3 + 1)

which simplifies to

C −D = 1

8. (11)

Substitution of x = 1 into (10) yields, after simplification,

3C +D =27

8. (12)

Solving equations (11) and (12) simultaneously yields that C = 7/8 andD = 3/4.Substitution of the values of A, B, C, D, and E which were found into

(8) yields the partial fraction decomposition

1

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

8

µ1

x− 8

x+ 1+

7

x+ 2+

6

(x+ 2)2+

4

(x+ 2)3

¶. (13)

Example 23 As our final example, let us write the rational expression

x4

x3 − 1in partial fraction form.

22

Since the degree of the polynomial in the numerator is not less than thedegree of the polynomial in the denominator, we must first perform long di-vision before attempting a partial fraction decomposition. After performingthe long division, we obtain

x4

x3 − 1 = x+x

x3 − 1 . (14)

The second term on the right hand side of (14) can be written in factoredform as

x

x3 − 1 =x

(x− 1)(x2 + x+ 1) . (15)

Since the factor x2+x+1 which appears in equation (15) is an irreduciblequadratic factor, we attempt to decompose (15) as a sum of the form

x

(x− 1)(x2 + x+ 1) =A

x− 1 +Bx+ C

x2 + x+ 1. (16)

After multiplication of both sides by the denominator of the left hand side,equation (16) becomes

x = A(x2 + x+ 1) + (Bx+ C)(x− 1). (17)

Substitutions of x = 1, x = 0, and x = −1 into (17) yield, respectively,that A = 1/3, C = 1/3, and B = −1/3.Going back to (14), we obtain the partial fraction decomposition

x4

x3 − 1 = x+1

3

µ1

x− 1 +−x+ 1x2 + x+ 1

¶.

Exercise 24 1. Obtain partial fraction decompositions of the followingrational expressions. Then evaluate the indefinite integral of each ex-pression (if you can).

(a)x− 2

x2 + 3x+ 2

(b)x− 2

x2 + 4x+ 4

23

(c)2

x(x2 − x+ 2)(d)

2

(x2 − 2x+ 1)(x2 + 3x+ 2)2

2. Use the results in Examples 22 and 23 to evaluate the indefinite integralsZ1

x(x+ 1)(x+ 2)3dx

and Zx4

x3 − 1 dx.

(You might find the second one to be challenging!)

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