Techniques of Integration - Whitman College

10

Techniques of Integration

10.1

Powers of sine and

osine

Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is

straightforward. Some examples will suffice to explain the approach.

EXAMPLE 10.1.1

Z

5

sin x dx =

Z

Evaluate

Z

4

sin5 x dx. Rewrite the function:

sin x sin x dx =

Z

2

2

sin x(sin x) dx =

Z

sin x(1 ? cos2 x)2 dx.

Now use u = cos x, du = ? sin x dx:

Z

2

2

sin x(1 ? cos x) dx =

=

Z

Z

?(1 ? u2 )2 du

?(1 ? 2u2 + u4 ) du

1

2

= ?u + u3 ? u5 + C

3

5

2

1

= ? cos x + cos3 x ? cos5 x + C.

3

5

203

204

Chapter 10 Techniques of Integration

EXAMPLE 10.1.2

Z

Evaluate

function:

sin6 x dx. Use sin2 x = (1 ? cos(2x))/2 to rewrite the

(1 ? cos 2x)3

dx

8

Z

1

=

1 ? 3 cos 2x + 3 cos2 2x ? cos3 2x dx.

8

Now we have four integrals to evaluate:

Z

1 dx = x

Z

6

sin x dx =

Z

2

3

(sin x) dx =

and

Z

Z

3

?3 cos 2x dx = ? sin 2x

2

are easy. The cos3 2x integral is like the previous example:

Z

Z

3

? cos 2x dx = ? cos 2x cos2 2x dx

Z

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

Z

1

= ? (1 ? u2 ) du

2





u3

1

u?

=?

2

3





1

sin3 2x

=?

sin 2x ?

.

2

3

And finally we use another trigonometric identity, cos2 x = (1 + cos(2x))/2:





Z

Z

sin 4x

1 + cos 4x

3

2

x+

.

3 cos 2x dx = 3

dx =

2

2

4

So at long last we get









Z

sin3 2x

3

sin 4x

3

1

x

6

sin 2x ?

+

x+

+ C.

sin 2x ?

sin x dx = ?

8 16

16

3

16

4

EXAMPLE 10.1.3 Evaluate

Z

sin2 x cos2 x dx. Use the formulas sin2 x = (1?cos(2x))/2

and cos2 x = (1 + cos(2x))/2 to get:

Z

Z

1 ? cos(2x) 1 + cos(2x)

2

2

sin x cos x dx =

¡¤

dx.

2

2

The remainder is left as an exercise.

10.2

Trigonometric Substitutions

205

Exercises 10.1.

Find Zthe antiderivatives.

2

sin x dx ?

1.

2.

Z

sin3 x dx ?

4.

Z

cos2 x sin3 x dx ?

6.

Z

sin2 x cos2 x dx ?

sin x(cos x)3/2 dx ?

tan3 x sec x dx ?

3.

Z

5.

Z

7.

Z

cos x sin x dx ?

8.

Z

9.

Z

sec2 x csc2 x dx ?

10.

Z

10.2

4

sin x dx ?

3

cos x dx ?

3

2

Trigonometri

Substitutions

So far we have seen that it sometimes helps to replace a subexpression of a function by

a single variable. Occasionally it can help to replace the original variable by something

more complicated. This seems like a ¡°reverse¡± substitution, but it is really no different in

principle than ordinary substitution.

EXAMPLE 10.2.1

Evaluate

Z p

1 ? x2 dx. Let x = sin u so dx = cos u du. Then

Z p

Z p

Z ¡Ì

2

2

1 ? x dx =

1 ? sin u cos u du =

cos2 u cos u du.

¡Ì

We would like to replace cos2 u by cos u, but this is valid only if cos u is positive, since

¡Ì

cos2 u is positive. Consider again the substitution x = sin u. We could just as well think

of this as u = arcsin x. If we do, then by the definition of the arcsine, ?¦Ð/2 ¡Ü u ¡Ü ¦Ð/2, so

cos u ¡Ý 0. Then we continue:

Z

Z

Z ¡Ì

u sin 2u

1 + cos 2u

2

2

du = +

+C

cos u cos u du = cos u du =

2

2

4

=

arcsin x sin(2 arcsin x)

+

+ C.

2

4

This is a perfectly good answer, though the term sin(2 arcsin x) is a bit unpleasant. It is

possible to simplify this. Using the identity sin 2x = 2 sin x cos x, we can write sin 2u =

q

p

p

2

2 sin u cos u = 2 sin(arcsin x) 1 ? sin u = 2x 1 ? sin2 (arcsin x) = 2x 1 ? x2 . Then the

full antiderivative is

¡Ì

¡Ì

arcsin x x 1 ? x2

arcsin x 2x 1 ? x2

+

=

+

+ C.

2

4

2

2

206

Chapter 10 Techniques of Integration

This type of substitution is usually indicated when the function you wish to integrate

contains a polynomial expression that might allow you to use the fundamental identity

sin2 x + cos2 x = 1 in one of three forms:

cos2 x = 1 ? sin2 x

sec2 x = 1 + tan2 x

tan2 x = sec2 x ? 1.

If your function contains 1 ? x2 , as in the example above, try x = sin u; if it contains 1 + x2

try x = tan u; and if it contains x2 ? 1, try x = sec u. Sometimes you will need to try

something a bit different to handle constants other than one.

EXAMPLE 10.2.2 Evaluate

Z p

4 ? 9x2 dx. We start by rewriting this so that it looks

more like the previous example:

Z p

Z p

Z p

2

2

4 ? 9x dx =

4(1 ? (3x/2) ) dx = 2 1 ? (3x/2)2 dx.

Now let 3x/2 = sin u so (3/2) dx = cos u du or dx = (2/3) cos u du. Then

Z p

Z

Z p

4

2

2

cos2 u du

2 1 ? (3x/2) dx = 2 1 ? sin u (2/3) cos u du =

3

4u 4 sin 2u

+

+C

=

6

12

2 arcsin(3x/2) 2 sin u cos u

+

+C

=

3

3

2 arcsin(3x/2) 2 sin(arcsin(3x/2)) cos(arcsin(3x/2))

=

+

+C

3

3

p

2 arcsin(3x/2) 2(3x/2) 1 ? (3x/2)2

=

+

+C

3

3

¡Ì

2 arcsin(3x/2) x 4 ? 9x2

=

+

+ C,

3

2

using some of the work from example 10.2.1.

EXAMPLE 10.2.3

Z p

1+

Evaluate

x2

Z p

1 + x2 dx. Let x = tan u, dx = sec2 u du, so

Z p

Z ¡Ì

2

2

dx =

1 + tan u sec u du =

sec2 u sec2 u du.

¡Ì

Since u = arctan(x), ?¦Ð/2 ¡Ü u ¡Ü ¦Ð/2 and sec u ¡Ý 0, so sec2 u = sec u. Then

Z ¡Ì

Z

2

sec2 u sec u du = sec3 u du.

In problems of this type, two integrals come up frequently:

Z

3

sec u du and

Both have relatively nice expressions but they are a bit tricky to discover.

Z

sec u du.

10.2

First we do

R

Trigonometric Substitutions

sec u du, which we will need to compute

Z

sec u du =

=

Z

Z

sec u

Z

207

sec3 u du:

sec u + tan u

du

sec u + tan u

sec2 u + sec u tan u

du.

sec u + tan u

Now let w = sec u + tan u, dw = sec u tan u + sec2 u du, exactly the numerator of the

function we are integrating. Thus

Z

Z

Z

1

sec2 u + sec u tan u

du =

dw = ln |w| + C

sec u du =

sec u + tan u

w

= ln | sec u + tan u| + C.

Now for

Z

sec3 u du:

sec3 u sec3 u

sec3 u (tan2 u + 1) sec u

sec u =

+

=

+

2

2

2

2

3

2

3

sec u sec u tan u sec u

sec u + sec u tan2 u sec u

=

+

+

=

+

.

2

2

2

2

2

3

We already know how to integrate sec u, so we just need the first quotient. This is ¡°simply¡±

a matter of recognizing the product rule in action:

Z

sec3 u + sec u tan2 u du = sec u tan u.

So putting these together we get

Z

sec u tan u ln | sec u + tan u|

+

+ C,

sec3 u du =

2

2

and reverting to the original variable x:

Z p

sec u tan u ln | sec u + tan u|

+

+C

1 + x2 dx =

2

2

sec(arctan x) tan(arctan x) ln | sec(arctan x) + tan(arctan x)|

+

+C

2

2

¡Ì

¡Ì

ln | 1 + x2 + x|

x 1 + x2

+

+ C,

=

2

2

q

p

using tan(arctan x) = x and sec(arctan x) = 1 + tan2 (arctan x) = 1 + x2 .

=

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