6.2 Trigonometric Integrals and Substitutions

Arkansas Tech University MATH 2924: Calculus II

Dr. Marcel B. Finan

6.2 Trigonometric Integrals and Substitutions

In this section, we discuss integrals with trigonometric integrands and integrals that can be transformed to trigonometric integrals by substitution.

Trigonometric Integrals In order to understand the following discussion, the reader is encouraged to review trigonometric identities before proceeding with the discussion.

Integrals of the form sinn xdx or cosn xdx

Example 6.2.1 Find sin3 xdx.

Solution. First, we notice that sin3 x = sin x sin2 x = sin x(1 - cos2 x). Using the

substitution u = cos x, where du = - sin xdx, we find

sin3 xdx = - (1 - u2)du = -u + u3 + C = - cos x + cos3 x + C

3

3

Example 6.2.2 Find cos4 xdx.

Solution. Using the trigonometric identity

cos2 x = 1 + cos 2x 2

1

we find

cos4 xdx = [cos2 x]2dx =

1 + cos 2x 2 dx

2

1 =

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

4

1 = dx + 2 cos 2xdx +

4

1 + cos 4x dx

2

1

sin 2x x 1 sin 4x

= x+2

++

4

2 2 24

13

sin 4x

= x + sin 2x +

42

8

Example 6.2.3 Find sin2 xdx.

Solution. Using the trigonometric identity

sin2 x = 1 - cos 2x 2

we find

sin2 xdx =

1 =

2 1 = 2

1 - cos 2x dx

2

(1 - cos 2x)dx

sin 2x

x-

+C

2

Integrals of the form sinn x cosm xdx

Example 6.2.4 Find sin3 x cos4 xdx.

2

Solution. Let u = cos x then du = - sin xdx. Thus,

sin3 x cos4 xdx = (1 - cos2 x) sin x cos4 xdx

= - (1 - u2)u4du = (-u4 + u6)du u5 u7

=- + +C 57

Example 6.2.5 Find sin2 x cos2 xdx.

cos5 x cos7 x

=-

+

+C

5

7

Solution. Using the trigonometric identity cos2 x + sin2 x = 1, we have

sin2 x cos2 xdx = sin2 x(1 - sin2 x)dx

= sin2 xdx - sin4 xdx

=

x sin 2x -

-

1 -

sin3

x

cos

x

+

3

sin2 xdx

24

4

4

x =

-

sin

2x

+

1

sin3

x

cos

x

-

3

x sin 2x -

+C

244

42 4

x =

-

sin

2x

+

1

sin3

x

cos

x

+

3

3 sin 2x - x + C

244

16

8

=-

1

sin

2x

+

1

sin3

x

cos

x

+

1 x

+

C

16

4

8

Integrals of Tangent and Secant

Example 6.2.6 Find tan xdx.

3

Solution. Using the substitution u = cos x, we find

sin x

du

tan xdx =

dx = -

cos x

u

= - ln |u| + C = - ln | cos x| + C

= ln | sec x| + C

Example 6.2.7 Find sec xdx.

Solution. First, note that

(sec x + tan x) = sec2 x + sec x tan x = sec x(sec x + tan x).

Using the substitution u = sec x + tan x, we find

sec xdx = =

sec x + tan x

sec x

dx

sec x + tan x

sec2 x + sec x tan x dx

sec x + tan x

du

=

= ln |u| + C

u

= ln | sec x + tan x| + C

4

Trigonometric Substitutions

Thissubsection deals with integrands involving terms like x2 - a2, x2 + a2,

and a2 - x2.

? Integrands involving a2 - x2, -a x a, a > 0.

For

each

x

in

the

interval

[-a,

a]

there

is

a

in

the

interval

[-

2

,

2

]

such

that

x=

a sin

(notice

that

-1

x a

1 and

recall

the

graph

of sin x).

Thus,

using

the

substitution

x

=

a

sin

,

-

2

2

to

obtain

a2 - x2 = a2(1 - sin2 )

= a2 cos2 = a| cos |

=a cos

where we have used the Pythagorean identity cos2 + sin2 = 1. Notice that

cos2

=

cos

since

cos

0

in

-

2

2

.

It

is

important

to

point

out

here that by constructing a right triangle with one of the angle being then

the and

hypotenuse of the the adjacent side

triangle haslength a, has length a2 - x2.

the opposite side has length x

It follows that cos =

a2-x2 a

.

See Figure 6.2.1.

Figure 6.2.1

Example 6.2.8

Find

1 4-x2

dx.

Solution.

Let

x

=

2

sin

,

-

2

<

<

2

.

Then

4 - x2 = 4 - 4 sin2 = 4 cos2 = 2 cos .

Moreover, dx = 2 cos d. It follows that

1

2 cos

x

dx =

d = + C = arcsin + C

4 - x2

2 cos

2

5

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