Section 7.3, Some Trigonometric Integrals

Section 7.3, Some Trigonometric Integrals

Homework: 7.3 #1¨C31 odds

We will look at five commonly encountered types of trigonometric integrals:

R

R

1. sinn x dx and cosn x dx

R

2. sinm x cosn x dx

R

R

R

3. sin mx cos nx dx, sin mx sin nx dx, and cos mx cos nx dx

R

R

4. tann x dx and cotn x dx

R

R

5. tanm x secn dx and cotm x cscn x dx

We will demonstrate how to calculate these by example. Throughout this section, we will be using

many trigonometric identities, including:

sin2 x + cos2 x = 1

1 + tan2 x = sec2 x

1 + cot2 x = csc2 x

1 ? cos 2x

sin2 x =

2

1

+

cos

2x

cos2 x =

2



1

sin mx cos nx =

sin(m + n)x + sin(m ? n)x

2



1

sin mx sin nx = ? cos(m + n)x ? cos(m ? n)x

2



1

cos mx cos nx =

cos(m + n)x + cos(m ? n)x

2

1

Integrals of the form

R

sinn x dx and

R

cosn x dx

We will look at examples when n is odd and when n is even. When n is odd, we will use sin2 x +

2x

2x

cos2 x = 1. When n is even, we will use either sin2 x = 1?cos

or cos2 x = 1+cos

.

2

2

Examples

R

1. Find cos5 x dx.

We will use the identity cos2 x = 1 ? sin2 x, so we will substitute cos4 x = (1 ? sin2 x)2 .

Z

Z

cos5 x dx = (1 ? sin2 x)2 cos x dx

Z




=

1 ? 2 sin2 x + sin4 x cos x dx

Z




=

cos x ? 2 sin2 x cos x + sin4 x cos x dx

= sin x ?

2

1

sin3 x + sin5 x + C

3

5

R

2. Find sin4 x dx

We will start by using sin2 x =

Z

Z 

4

sin x dx =

1?cos 2x

.

2

1 ? cos 2x

2

2

dx

Z

=

=

=

=

=




1

1 ? 2 cos 2x + cos2 2x dx

4

Z

Z

Z

1

1

1

dx ?

cos 2x dx +

cos2 2x dx

4

2

4

Z

x 1

1

1 + cos 4x

? sin 2x +

dx

4 4

4

2

x

1

x 1

? sin 2x + +

sin 4x + C

4 4

8 32

3x 1

1

? sin 2x +

sin 4x + C,

8

4

32

where we also used that cos2 x =

2

Integrals of the form

1+cos 2x

2

R

in the third-to-last line.

sinm x cosn x dx

If either m or n is an odd positive integer, we will use the identity sin2 x + cos2 x = 1. If both m

and n are even and positive, we will use the half-angle identities.

Examples

R

1. Find cos5 x sin?4 x dx

Since the exponent for cosine is odd, we will replace cos4 x by (1?sin2 x)2 = 1?2 sin2 x+sin4 x:

Z

Z

?4

5

cos x sin x dx = cos x(1 ? 2 sin2 x + sin4 x) sin?4 x dx

Z

Z

Z

= cos x sin?4 x dx ? 2 cos x sin?2 x dx + cos x dx

1

= ? (sin x)?3 + 2(sin x)?1 + sin x + C

3

(Be careful with notation, since sin?1 x refers to the inverse sine function, not 1/(sin x).)

R

2. Find cos2 x sin4 x dx.



2

4

1+cos 2x

1?cos 2x

2

We will substitute cos x =

and sin x =

. Then,

2

2

Z

cos2 x sin4 x dx =

=

=

=

=

=

=



2

1 + cos 2x 1 ? cos 2x

dx

2

2

Z

1

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

8

Z

1

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

8

Z

Z

Z

Z

1

1

1

1

2

dx ?

cos 2x dx ?

cos 2x dx +

cos3 2x dx

8

8

8

8

Z

Z

x

1

1

1 + cos 4x

1

?

sin 2x ?

dx +

cos 2x(1 ? sin2 2x) dx

8 16

8

2

8

x

1

x

1

1

1

?

sin 2x ?

?

sin 4x +

sin 2x ?

sin3 2x + C

8 16

16 64

16

48

x

1

1

?

sin 4x ?

sin3 2x + C

16 64

48

Z

3

RIntegrals of the form

cos mx cos nx dx

R

sin mx cos nx dx,

R

sin mx sin nx dx, and





1

Here,

we

will

use

the

identities

sin

mx

cos

nx

=

sin(m

+

n)x

+

sin(m

?

n)x

, sin mx sin nx =

2







1

1

? 2 cos(m + n)x ? cos(m ? n)x , and cos mx cos nx = 2 cos(m + n)x + cos(m ? n)x .

Examples

R

1. Find sin 4x cos 5x dx 



Since sin mx cos nx = 21 sin(m + n)x + sin(m ? n)x , we will use this with m = 4 and n = 5:

Z

Z




1

sin 9x + sin(?x) dx

sin 4x cos 5x dx =

2

1

1

= ? cos 9x + cos x + C,

18

2

where we used that cos(?x) = cos x.

R¦Ð

2. Find ?¦Ð sin mx sin nx dx, where m and n are positive integers.

First, consider m 6= n. Then,

Z ¦Ð

Z



1 ¦Ð 

sin mx sin nx dx = ?

cos(m + n)x ? cos(m ? n)x dx

2

?¦Ð

?¦Ð





1

1

1

sin(m + n)x ?

sin(m ? n)

=?

2 m+n

m?n

¦Ð

?¦Ð

=0

If m = n, then

Z ¦Ð

Z



1 ¦Ð 

sin mx sin nx dx = ?

cos(2m)x ? 1 dx

2 ?¦Ð

?¦Ð



¦Ð

1 1

=?

sin(2m)x ? x

2 2m

?¦Ð

=¦Ð

4

Integrals of the form

R

tann x dx and

R

cotn x dx

In the tangent case, we will use tan2 x = sec2 x ? 1. In the cotangent case, we will use cot2 x =

csc2 x ? 1. Here, we will only replace tan2 x or cot2 x, distribute, integrate what we can, then repeat

as necessary.

Examples

1. Find

R

Z

tan4 x dx

tan4 x dx =

Z

tan2 x(sec2 x ? 1) dx

Z

Z

2

2

= tan x sec x dx ? tan2 x dx

Z

1

= tan3 x ? (sec2 x ? 1) dx

3

1

= tan3 x ? tan x + x + C

3

2. Find

R

Z

5

cot5 x dx

cot5 x dx =

Z

cot3 x(csc2 x ? 1) dx

Z

Z

3

2

= cot x csc x ? cot3 x dx

Z

1

= ? cot4 x ? cot x(csc2 x ? 1) dx

4

Z

Z

1

cos x

= ? cot4 x ? cot x csc2 x dx +

dx

4

sin x

1

1

= ? cot4 x + cot2 x + ln | sin x| + C

4

2

Integrals of the form

R

tanm x secn dx and

R

cotm x cscn x dx

If n is even, use either sec2 x = tan2 x + 1 or csc2 x = cot2 x + 1 to replace all but 2 powers of sec x

or csc x, then you can use a u-substitution to integrate.

If m is odd, we will use that the derivative of sec x is sec x tan x (or the derivative of csc x is

? csc x cot x), and replace m ? 1 powers of either tangent or cotangent using a Pythagorean identity.

Examples

1. Find

R

Z

tan1/2 x sec4 x dx

tan1/2 x sec4 x dx =

tan1/2 x(tan2 +1) sec2 x dx

Z

Z

= tan5/2 x sec2 x dx + tan1/2 x sec2 x dx

=

2. Find

R

Z

Z

2

2

tan7/2 x + tan3/2 x + C

7

3

cot3 x csc3/2 x dx

cot3 x csc3/2 x dx =

Z

cot x(csc2 x ? 1) csc3/2 x dx

Z

Z

= cot x csc x csc5/2 x dx ? cot x csc x csc1/2 x dx

2

2

= ? csc7/2 x + csc3/2 +C

7

3

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