Trigonometric Integrals - Lia Vas

Calculus 2 Lia Vas

Trigonometric Integrals

Let us consider the integrals of the form

f (sin x) cos xdx

or

f (cos x) sin xdx

where f (x) is a function with antiderivative F (x). Using the substitution

u = sin x for the first integral and

u = cos x for the second integral

one can reduce the first integral to f (u)du = F (u) + c = F (sin x) + c and the second integral to f (u)(-du) = -F (u) + c = -F (cos x) + c.

This idea can be applied to the integrals of the form

sinn x cosm x dx

if either m or n are odd. We will refer to this situation as the good case. If n is odd rewrite sinn-1 x as a function of cos x using the trigonometric identity sin2 x =

1-cos2 x. Note that then you obtain an integral of the form f (cos x) sin xdx where f is a polynomial function (thus easy to integrate). You can evaluate the integral using the substitution

u = cos x.

If m is odd rewrite cosm-1 x as a function of sin x using the trigonometric identity cos2 x = 1-sin2 x. Note that then you obtain an integral of the form f (sin x) cos xdx where f is a polynomial function. You can evaluate the integral using the substitution

u = sin x.

Let us consider the case when both n or m are even. Let us refer to this as the bad case (although it is not that bad). The idea is to use one or more of the following three trigonometric identities

sin2 x

=

1 (1

-

cos 2x)

2

cos2

x

=

1 (1

+

cos 2x)

and

1 sin x cos x = sin 2x

2

2

to reduce the integral to a sum of integrals in which the powers of cosines and sines are at most 1.

Then you can integrate term by term.

If you encounter a multiple of x in the argument of sin or cos, note that the three identities above

become

sin2 ax

=

1 2

(1

-

cos 2ax),

cos2 ax

=

1 2

(1

+ cos 2ax),

and

sin ax

cos ax

=

1 2

sin 2ax.

Note: If you have to integrate other trigonometric functions, you can convert them to sin and cos

functions using the trigonometric identities.

1

Practice Problems. Evaluate the following integrals:

1. sin10 x cos x dx

2. sin3 x cos2 x dx

3. ecos x sin x dx

cos x 4. 1 + sin2 x dx

5. tan x dx

6. cos2 x tan x dx

7. sin2 x dx

8. sin2 x cos2 x dx

9. sin5 x dx

10. cos4 x dx

11. Finding the center of mass. Let R be the region between the graphs of f and g such that f (x) g(x) on interval [a, b]. The area A of R is A = ab(f (x) - g(x)) dx. Then the center of mass of R is the point (x?, y?) where

1b

x? =

x (f (x) - g(x)) dx

Aa

1 y? =

b 1 ((f (x))2 - (g(x))2) dx

A a2

Find the center of mass of the region bounded by the given curves. (a) y = sin x, y = cos x, x = 0, x = /4. (b) y = sin 2x, y = 0, x = 0, x = /2.

Solutions.

1. This integral is already in the form f (sin x) cos xdx so use the substitution u = sin x. The

integral becomes

u10

cos

x

du cos x

=

u10du

=

1 11

u11

+c

=

1 11

sin11 x

+ c.

2. The sine function has the odd power. So, this is the "good case". Write sin3 x as sin2 x sin x =

(1 - cos2 x) sin x and get cos2 x sin3 x dx = cos2 x (1 - cos2 x) sin x dx = (cos2 x -

cos4 x) sin x dx. This integral is of the form f (cos x) sin x dx that can be evaluated using

u

=

cos x

du

=

- sin xdx

du - sin x

=

dx.

The

integral

reduces

to

(cos2 x-cos4 x) sin x dx =

(u2

-

u4)

sin

x

-

du sin

x

=

-

(u2

- u4)du

=

-1 3

u3

+

1 5

u5

+

c

=

-1 3

cos3

x

+

1 5

cos5 x

+ c.

3. This integral is of the form f (cos x) sin xdx so use the substitution u = cos x. The integral

becomes

eu

sin

x

-

du sin

x

=

-

eudu = -eu + c = -ecos x + c.

4. This integral is of the form f (sin x) cos xdx so use the substitution u = sin x. The integral

becomes

cos 1+u2

du cos x

=

1 1+u2

du

=

tan-1

u

+

c

=

tan-1(sin x) +

c.

5.

Write tan x as

sin x cos x

=

1 cos

x

sin

x

and

treat

this

as

f

(cos

x)

sin

x.

So,

use

the

substitution

u

=

cos

x

to obtain

1 u

sin

x

-

du sin

x

=

-

1 u

du

=

-

ln

|u|

+

c

=

-

ln

|

cos

x|

+

c.

2

6.

cos2 x tan x dx =

cos2

x

sin cos

x x

dx

=

cos x sin x. Note that this is a good case with both

exponents equal (super good case!) so both substitutions u = cos x and u = sin x could work.

With u = sin x, one obtains

cos

xu

du cos x

=

1 2

u2

+c

=

1 2

sin2 x +

c.

With u = cos x, one obtains

u

sin

x

-

du sin

x

=

-1 2

u2

+

c

=

-1 2

cos2

x + c.

Note

that

this

is

the

same

as

the

previous

answer

since

-1 2

cos2

x+c

=

-1 2

(1

-

sin2

x) + c

=

-1 2

+

1 2

sin2 x + c

=

1 2

sin2 x + c1.

7. This is the bad case.

Use the trigonometric identity sin2 x =

1 2

(1

-

cos 2x)

to

have

1 2

(1

-

cos 2x)dx.

Use

the

substitution

u

=

2x

and

obtain

1 2

x

-

1 4

sin(2x) + c.

8.

This is the bad case as well.

Use both identities sin2 x =

1 2

(1 - cos

2x)

and

cos2

x

=

1 2

(1

+

cos

2x)

to have

sin2 x cos2 xdx =

1 4

(1

-

cos

2x)(1

+

cos

2x)dx

=

1 4

(1

-

cos2

2x)dx.

Then

use

the

trig

identity

cos2 x

=

1 2

(1

+

cos

2x)

with

2x

instead

of

x

to

reduce

cos2 2x

to

linear

terms.

Obtain

1 4

(1

-

1 2

(1

+

cos

4x))dx

=

(

1 4

-

1 8

-

1 8

cos

4x)dx

=

(

1 8

-

1 8

cos 4x)dx

=

1 8

x

-

1 32

sin 4x

+

c.

Alternatively,

you

can

use

sin x cos x

=

1 2

sin 2x

first

and

then

sin2

2x

1 2

(1

-

cos

4x)

after.

In

this

case, you will get (sin x cos x)2dx =

1 4

sin2

2xdx

=

1 8

(1

-

cos 4x)dx

=

1 8

x

-

1 32

sin 4x

+

c.

9. This is the good case. sin5 x = sin4 x sin x = (sin2 x)2 sin x = (1 - cos2 x)2 sin x. Since this

is of the form f (cos x) sin x, you can use the substitution u = cos x. The integral becomes

(1 - u2)2

sin

x

-

du sin

x

=

-(1

-

2u2

+

u4)du

=

-u

+

2 3

u3

-

1 5

u5

=

-

cos

x

+

2 3

cos3

x

-

1 5

cos5

x

+ c.

10.

This is the bad case.

Using

the

trig

identities

cos4 x

=

(cos2 x)2

=

(

1 2

(1

+

cos 2x))2

=

1 4

(1

+

2 cos 2x + cos2

2x)

=

1 4

(1

+

2

cos

2x

+

1 2

(1

+

cos

4x))

=

1 4

+

1 2

cos 2x +

1 8

+

1 8

cos 4x

=

3 8

+

1 2

cos 2x +

1 8

cos 4x.

Integrating

term

by

term

now,

you

obtain

3 8

x

+

1 4

sin 2x +

1 32

sin 4x + c.

11.

(a)

Note

that

on

(0,

4

)

0/4(cos x - sin x)dx =

cos x is larger than sin x. So the

(sin x + cos x)|0/4

=

2-1

area A can be evaluated as .414. The x-coordinate x? is

A x?

= =

1 .414

this

/4 0

x(cos

x-sin

x)

integral.

x? =

1 .414

dx. Use x(sin x

the integration by parts with u = x and dv =

+ cos x)|0/4 - (- cos x + sin x)|0/4 dx

=

1 .414

(cos x-sin

(

4

2 - 1)

x)dx for = .267.

y? =

1 .414

/4 0

1 2

(cos2

x

-

sin2 x)dx.

This

is

the

"bad

case".

Using the trigonometric identities

for

sin2 x

and

cos2 x,

we

obtain

1 .414

1 4

0/4(1 + cos 2x - 1 + cos 2x)dx

=

1 1.656

/4 0

2

cos

2xdx

=

1 1.656

sin

2x|0 /4

=

.6035.

So,

the

center

of

mass

is

(.267,

.6035).

(b) A =

/2 0

sin

2xdx

=

-1 2

cos

2x|0/2

=

-1 2

(-1

-

1)

=

1.

x?

=

/2 0

x

sin

2xdx.

Using

the

integration

by

parts

obtain

that

x?

=

(

-x 2

cos 2x

+

1 4

sin 2x)|0/2

=

4

.

y?

=

/2 0

1 2

sin2

2xdx.

Using

the

trigonometric

identities

for

sin2 x,

we

obtain

1 4

0/2(1 - cos 4x)dx

=

1 4

(x

-

1 4

sin 4x)|0/2

=

8

.

So,

the

center

of

mass

is

(

4

,

8

).

3

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download