Trigonometric Integrals

[Pages:5]Trigonometric Integrals

Liming Pang

When dealing with integrals involving trigonometric functions, the following formulas from precalculus are often helpful:

1. sin2 x + cos2 x = 1, 1 + tan2 x = sec2 x, 1 + cot2 x = csc2 x

2.

sin x =

1 csc

x

,

cos x =

1 sec

x

,

tan

x

=

1 cot x

3.

tan x =

sin cos

x x

,

cot x

=

cos x sin x

4. sin 2x = 2 sin x cos x, cos 2x = cos2 x-sin2 x = 1-2 sin2 x = 2 cos2 x-1

Example 1. Evaluate cos3 x dx

cos3 x dx = cos2 x d sin x

= 1 - sin2 x d sin x

= sin x - 1 sin3 x + C 3

Example 2. Evaluate

0

sin2

x

dx

sin2 x dx =

1 - cos 2x dx

0

0

2

1 2 1 - cos u

=

du

20

2

1 2

2

= ( 1 du - cos u du)

40

0

=

2

1

Example 3. Evaluate sin3 x cos2 x dx

sin3 x cos2 x dx = - sin2 x cos2 x d cos x

= - (1 - cos2) cos2 x d cos x

= cos4 x - cos2 x d cos x

= 1 cos5 x - 1 cos3 x + C

5

3

Example 4. Evaluate cos4 x dx

cos4 x dx = ( 1 + cos 2x )2 dx 2

1 =

cos2 2x + 2 cos 2x + 1 dx

4

1 1 + cos 4x

=

+ 2 cos 2x + 1 dx

4

2

11

3

=

cos 4x + 2 cos 2x + dx

42

2

1

1

3

= cos 4x dx + cos 2x dx + 1 dx

8

2

8

1

1

3

=

cos 4x d4x + cos 2x d2x + x

32

4

8

1

1

3

= sin 4x + sin 2x + x + C

32

4

8

Sometimes we will make use of trigonometric functions to do some integration by substitution. Some general idea is the following:

a2

- x2

let

x

=

a sin t, -

t

a2

+

x2

let

x

=

2 a tan t, -

<

t

<

2

2

2

x2 - a2

let

x = a sec t, 0 t <

or

t<

3

2

2

2

Example 5. Evaluate

4 - x2 dx

Let

x

=

2 sin t,

where

-

2

t

2

.

4 - x2 dx =

4 - 4 sin2 t d2 sin t

= 2 cos t d2 sin t = 4 cos2 dt

= (1 + cos 2t) d2t

= 2t + sin 2t + C

=

2

sin-1

x

+

x 4

-

x2

+

C

22

Example 6. Find the area enclosed by the ellipse

x2 y2 + =1

a2 b2

When

y

0,

y

=

b a

a2 - x2, so the area is

a b

2b a

2

a2 - x2 dx =

a2 - x2 dx

-a a

a -a

Let

x

=

a sin t,

-

2

t

2

,

then

2b

a

2b

a2 - x2 dx =

2

a2 - a2 sin2 t da sin t

a -a

a-

2

= 2ab 2 cos2 t dt

-

2

2 1 + cos 2t

= 2ab

dt

-

2

2

2

1

= ab( 1 dt +

cos u du)

-

2

2 -

= ab

3

Example 7. Find

1 x2 x2+4

dx

Let

x

=

2 tan t,

-

2

<

t

<

2

,

then

1

1

dx =

x2 x2 + 4

(4 tan2 t)2 sec t d2 tan t

1 cos2 t

1

= 4

sin2 t ? cos t ? cos2 t dt

1 cos t

= 4

sin2 t dt

11

= 4

sin2 t d sin t

1

=-

+C

4 sin t

4 + x2

=-

+C

4x

The

last

equality

follows

from

1 sin t

= sec t =

1 + cot2 t =

Example 8. Evaluate

dx 3

10

x2

0 9-25x2

1

+

4 x2

Let

x=

3 5

sin

t,

when

x

is

from

0

to

3 10

,

t

is

from

0

to

6

3 10

x2

6

dx =

0 9 - 25x2

0

(

3 5

sin

t)2

3 d sin t

9 - 9 sin2 t 5

= 6 9 sin2 t dt

0 125

9 6 1 - cos 2t

=

d2t

125 0

4

9 3 1 - cos u

=

du

125 0

4

9 3

= (- )

500 3 2

3 9 3

=-

500 1000

4

Example 9. Compute

1 x2-a2

Let

x = a sec t,

0 ................
................

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