Section 7.2 Advanced Integration Techniques: Trigonometric ...

[Pages:6]Section 7.2

Advanced Integration Techniques: Trigonometric Integrals

When attempting to evaluate integrals of trig functions, it often helps to rewrite the function of interest using an identity. Thus we will use the following identities quite often in this section; you would do well to memorize them.

sin2(x)

=

1-cos(2x) 2

cos2(x)

=

1+cos(2x) 2

(1)

cos(2x) = 1 - 2 sin2 x

cos(2x) = 2 cos2 x - 1

sec2 x = 1 + tan2 x

csc2 x = 1 + cot2 x

There are many different possibilities for choosing an integration technique for an integral involving trigonometric functions. For example, we can solve

sin x cos x dx

using the u-substitution u = cos x. The same substitution could be used to find

tan x dx

if

we

note

that

tan x

=

sin cos

x x

.

We

can

use

integration

by

parts

to

solve

sin(5x) cos(3x) dx.

However, there are many other trigonometric functions whose integrals can not be evaluated so easily. In this section, we will look at multiple techniques for handling integrals of several different types of trig functions.

Integrals of the form sinm x cosn x

To integrate a function of the form sinm x cosn x dx,

which is a product of (positive integer) powers of sin x and cos x, we will use one of the two following methods:

1. If both the powers m and n are even, rewrite both trig functions using the identities in (1). 2. If at least one of the powers is odd, we will rewrite the original function so that only one power

of sin x (or one power of cos x) appears; this will allow us to make a helpful substitution: (a) If m = 2k + 1 is odd, then rewrite

sinm x = sin2k+1 x = (sin x)(sin2k x) = (sin x)(sin2 x)k = (sin x)(1 - cos2 x)k, and use the u-substitution u = cos x.

1

Section 7.2

(b) If n = 2k + 1 is odd, then rewrite cosn x = cos2k+1 x = (cos x)(cos2k x) = (cos x)(cos2 x)k = (cos x)(1 - sin2 x)k,

and use the u-substitution u = sin x. Note: The general idea behind this technique actually works for any integral of the form

sinm(x) cosn(x),

where is any real number.

Example. Find cos3(2x) dx.

Since cos(2x) has an odd power, let's rewrite cos3(2x) = cos(2x) cos2(2x) = cos(2x)(1 - sin2(2x)).

Then

cos3(2x)dx = cos(2x)(1 - sin2(2x)) dx.

We will need the substitution u = sin(2x) so that du = 2 cos(2x) dx. Now we can finish the problem:

cos3(2x) dx = cos(2x)(1 - sin2(2x)) dx

1 =

1 - u2 du

2

1 =

u - 1 u3

+C

2

3

=

1 u

-

1 u3

+

C

26

= 1 sin(2x) - 1 sin3(2x) + C.

2

6

using the substitution u = sin(2x)

Example. Find sin3 x cos5 xdx.

Since both trig functions have odd powers, we will rewrite one of them using the Pythagorean identity. Let's try

sin3 x cos5 x = sin3 x cos4 x cos x = sin3 x(cos2 x)2 cos x = sin3 x(1 - sin2 x)2 cos x.

2

Section 7.2

As in the previous example, we can use a simple u-substitution to finish the problem. Set

u = sin x so that du = cos x dx.

Then

sin3 x(1 - sin2 x)2 cos x dx = u3(1 - u2)2 du

= u3(1 - 2u2 + u4) du

= u3 - 2u5 + u7 du

= 1 u4 - 2 u6 + 1 u8 + C 468

= 1 sin4 x - 1 sin6 x + 1 sin8 x + C.

4

3

8

Example. Find cos2(2x)dx.

Since there are no odd powers in this function, we will rewrite the integrand as

cos2(2x) = 1 + cos(4x) 2

using the equation in (1). Then the integral calculation is fairly routine:

cos2(2x) dx =

1 + cos(4x) dx

2

1 = 1 + cos(4x) dx

2

11 = (x + sin(4x)) + C

24 11 = x + sin(4x) + C. 28

using the substitution u = 4x

Example. Evaluate cos2 x sin4 x dx. Since both the powers of cos x and sin x are even, we will write cos2 x = 1 + cos(2x) 2

and

2

sin4 x = (sin2 x)2 = 1 - cos(2x) . 2

3

Section 7.2

Then

cos2 x sin4 x dx =

1 + cos(2x) 2

2

1 - cos(2x) dx

2

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

=

dx

2

4

1 =

1 - 2 cos(2x) + cos2(2x) + cos(2x) - 2 cos2(2x) + cos3(2x) dx

8

1 =

1 - cos(2x) - cos2(2x) + cos3(2x) dx

8

1

1

= x - sin 2x -

cos2(2x) dx +

cos3(2x) dx.

8

2

We have already showed that

cos2(2x)

dx

=

1 x

+

1

sin(4x)

+

C

28

and so finally we have

cos3(2x) dx = 1 sin(2x) - 1 sin3(2x) + C,

2

6

cos2 x sin4 x dx = 1

x

-

1

sin

2x

-

1 x

-

1

sin(4x)

+

1

sin(2x)

-

1

sin3(2x)

+ C.

8

2

28

2

6

Integrating powers of tan x, sec x, csc x, and cot x

To integrate powers of the other trig functions, we will often need to use u-substitution or integration

by parts together with the pythagorean identities; if possible, we will need to take advantage of the

fact that

d tan x = sec2 x, d sec2 x = sec x tan x,

dx

dx

d csc x = - csc x cot x, and d cot x = - csc2 x.

dx

dx

Example. Evaluate csc4 xdx.

Rewriting

csc4 x = (csc2 x)(csc2 x) = (1 + cot2 x)(csc2 x)

4

Section 7.2

is advantageous, as it will allow us to use the substitution u = cot x:

csc4 x dx = (1 + cot2 x)(csc2 x) dx

= - (1 + u2) du = -u - 1 u3 + C

3 = - cot x - 1 cot3 x + C.

3

using u = cot x and - du = csc2 x dx

Eliminating Square Roots

If the function we wish to integrate involves the square root of some trigonometric function, we may be able to eliminate the root by using the pythagorean identities or the identities from (1).

Example. Evaluate cos y + 1 dy.

The identity can help us here. We have

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

cos2 y

1 + cos y

=

.

2

2

We would like to replace the quantity cos y + 1; solving for this expression in the above identity,

we have cos y + 1 = 2 cos2

y 2

. So we may rewrite the integral as

cos(y) + 1 dy =

2 cos2 y dy 2

y

= 2 cos dy

2

y

= 2 2 sin + C.

2

Example. Find csc2 - 1 d.

5

Section 7.2

Since csc2 - 1 = cot2 , let's rewrite

csc2 - 1 d =

cot2 d

= cot d.

cos Since cot = , we can integrate the function using a substitution; setting u = sin so that

sin du = cos d, we have

csc2 - 1 d = cot d

cos

=

d

sin

1 = du

u

= ln u + C

= ln | sin | + C.

A brief aside We have not yet learned how to evaluate sec x dx, and as we will need to know this integral

in future sections, let's go ahead and compute it. It turns out that the best way to evaluate the integral is by using Mathemagic: note that sec x can be rewritten as

sec x + tan x sec2 x + sec x tan x

sec x = sec x

=

.

sec x + tan x

sec x + tan x

This may seem pointless, but it will actually allow us to use a basic substitution to evaluate the integral. Setting u = sec x + tan x so that du = sec x tan x + sec2 x, we have

sec xdx = =

sec2 x + sec x tan x dx

sec x + tan x 1

du u

= ln |u| + C

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

6

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