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Trigonometric Integrals 1

Lecture Notes

page 1

Sample Problems

Compute each of the following integrals. Assume that a and b are positive numbers.

1.

Z

2.

Z

sin x dx

cos 5x dx

3.

Z

4.

Z

csc2 x

5.

Z

tan x dx

6.

Z

7.

Z

cos x sin4 x dx

dx

cot x dx

sec x dx

8.

Z

csc x dx

9.

Z

sin2 x dx

10.

Z

sin3 x dx

11.

Z

sin x dx

12.

Z

sin5 x dx

13.

Z

1

dx

2

a + b2 x2

14.

Z

15.

Z

p

sec ( x)

p

dx

x

16.

Z=3

p

1 + cos 2x dx

Z=2

p

1

0

17.

4

cos x dx

0

p

1

a2

x2

18.

dx

Z

tan3 x dx

19.

Z

sin 7x cos 3x dx

20.

Z

sin 10x sin 4x dx

15.

Z=6

Practice Problems

1.

2.

3.

4.

5.

6.

7.

Z

Z

Z

Z

Z

Z

Z

cos 3x dx

sin 4x

8.

5

dx

sec tan d

sec2

d

x tan x2

cot (2x

cos2 x

9.

10.

11.

dx

) dx

dx

c copyright Hidegkuti, Powell, 2012

12.

13.

14.

Z

Z

Z

Z

Z

Z

Z

cos2 (2x) dx

16.

sin 2a cos 8a da

17.

Z

cos b cos 11b db

18.

Z

sin 6 sin 14 d

19.

Z

cos 11m sin 3m dm

sin x cos5 x dx

tan2 x dx

cos 6x dx

Z

cos5 x dx

sin3 x cos5 x dx

1

0

cos3 x dx

cos4 x dx

p

Last revised: December 8, 2013

Trigonometric Integrals 1

Lecture Notes

page 2

Sample Problems - Answers

1.)

cos x + C

1

sin 5x + C

5

2.)

6.) ln jsin xj + C

3.)

7.) ln jsec x + tan xj + C

10.)

1

cos3 x

3

cos x + C

11.)

13.)

1

tan

ab

b

x +C

a

14.) sin

1

p

17.) 2 2 2

1

sin5 x + C

5

18.)

8.)

4.)

cot x + C

ln jcsc x + cot xj + C

1

1

3

sin 2x +

sin 4x + x + C

4

32

8

1

12.)

ln jcos xj + C = ln jsec xj + C

1

x

2

9.)

cos x +

1

sin 2x + C

4

2

cos3 x

3

p

p

15.) 2 ln jsec ( x) + tan ( x)j + C

x

+C

a

1

sec2 x + ln jcos xj + C

2

5.)

19.)

1

cos 10x

20

1

cos 4x + C

8

20.)

1

cos5 x + C

5

p

6

16.)

2

1

sin 6x

12

1

sin 14x + C

28

Practice Problems - Answers

1.)

1

sin 3x + C

3

2.)

6.)

1

ln jsin (2x

2

)j + C

10.)

1

cos 4x

4

5

3

1

1

x + sin 2x +

sin 4x + C

8

4

32

1

1

cos6 x + cos8 x + C

6

8

17.)

1

1

sin 10b +

sin 12b + C

20

24

3.) sec + C

1

1

x + sin 2x + C

2

4

7.)

13.)

+C

c copyright Hidegkuti, Powell, 2012

18.)

1

1

x + sin 4x + C

2

8

5.)

1

ln sec x2

2

9.) sin x

+C

1

sin3 x + C

3

2

1

sin3 x + sin x + C

12.)

cos6 x + C

3

6

p

2

1

1

x + tan x + C

15.)

16.)

cos 6a

cos 10a + C

3

12

20

11.)

14.)

8.)

4.) tan + C

1

sin5 x

5

1

sin 8

16

1

sin 20 + C

40

19.)

1

cos 8m

16

1

cos 14m + C

28

Last revised: December 8, 2013

Trigonometric Integrals 1

Lecture Notes

page 3

Sample Problems - Solutions

1.

Z

sin x dx

Solution: This is a basic integral we know from di��erentiating

basic trigonometric functions. Since

Z

d

d

cos x = sin x, clearly

( cos x) = sin x and so

sin x dx = cos x + C .

dx

dx

Z

2.

cos 5x dx

d

Solution: We know that

cos x = sin x + C. We will use substitution. Let u = 5x and then du = 5dx

dx

du

and so

= dx.

5

Z

Z

Z

du

1

1

cos 5x dx = cos u

=

cos u du = sin 5x + C

5

5

5

Note: Once we

Z have enough practice, there is no need to perform this substitution in writing. We can just

1

simply write

cos 5x dx = sin 5x + C.

5

Z

3.

cos x sin4 x dx

Solution: Let u = sin x. Then du = cos xdx.

Z

Z

Z

1

1

cos x sin4 x dx = sin4 x (cos xdx) = u4 u = u5 + C = sin5 x + C

5

5

4.

Z

csc2 x dx

d

cot x =

dx

Z

Z

2

csc x dx =

Solution: We need to remember that

5.

Z

csc2 x.

csc2 x dx =

tan x dx

Solution: Let u = cos x. Then du = sin x dx.

Z

Z

Z

Z

1

1

sin x

tan x dx =

dx =

(sin xdu) =

( du) =

cos x

u

u

= ln (cos x)

6.

Z

cot x + C

1

Z

1

du =

u

ln juj + C =

ln jcos xj + C

+ C = ln jsec xj + C

cot x dx

Solution: Let u = sin x. Then du = cos x dx.

Z

Z

Z

Z

cos x

1

1

cot x dx =

dx =

(cos xdu) =

du = ln juj + C = ln jsin xj + C

sin x

u

u

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

Trigonometric Integrals 1

Lecture Notes

7.

Z

sec x dx

Z

Solution:

sec x dx =

Z

sec2 x

Z

sec2 x + sec x tan x

dx =

sec x + tan x

Z

1

u

dx.

2

sec x + sec x tan x dx =

Z

csc x dx =

Z

Z

1

du = ln juj + C = ln jsec x + tan xj + C

u

csc2 x + csc x cot x

dx =

csc x + cot x

=

Z

csc2 x + csc x cot x

dx

csc x + cot x

d

csc x =

dx

csc x cot x dx.

Recall that

csc2 x

u = csc x + cot x. Then du =

Z

Z

csc x + cot x

dx =

csc x + cot x

csc x

From here we will use substitution.

9.

1

u

csc2 x + csc x cot x dx =

ln jcsc x + cot xj + C

1

(1

2

cos 2x)

1 dx

Z

sin2 x =

Z

2

sin x dx =

=

10.

csc x cot x and

d

cot x =

dx

csc2 x.

Z

Z

ln juj + C

1

( du) =

u

1

du =

u

Let

sin2 x dx

Solution: Recall the double angle formula for cosine, cos 2x = 1

Z

Let u =

csc x dx

Solution:

Z

sec2 x + sec x tan x

dx

sec x + tan x

d

d

sec x = sec x tan x and

tan x = sec2 x.

dx

dx

Recall that

sec x + tan x. Then du = sec x tan x +

Z

Z

sec x + tan x

sec x

dx =

sec x + tan x

From here we will use substitution.

8.

page 4

Z

1

(1

2

1

cos 2x) dx =

2

1

sin 2x + C

4

1

x

2

Z

2 sin2 x. We solve this for sin2 x

cos 2x dx

=

1

2

x + C1

1

sin 2x + C2

2

sin3 x dx

Solution:

Z

3

sin x dx =

Z

2

sin x sin x dx =

Let u = cos x: Then du = sin xdx

Z

Z

Z

3

2

sin x dx =

sin x 1 cos x dx =

=

1 3

u

3

u+C =

c copyright Hidegkuti, Powell, 2012

1

cos3 x

3

1

2

Z

sin x 1

cos x (sin xdx) =

cos2 x dx

Z

1

u

2

( du) =

Z

u2

1 du

cos x + C

Last revised: December 8, 2013

Trigonometric Integrals 1

Lecture Notes

11.

Z

page 5

sin4 x dx

Solution: We use the double angle formula for cosine to express sin2 x.

2 sin2 x

cos 2x = 1

Z

4

sin x dx =

Z

2

sin x

2

dx =

Z

=)

2

1

(1

2

cos 2x)

sin2 x =

1

dx =

4

Z

1

(1

2

(1

cos 2x)

1

cos 2x) dx =

4

2

We use the double angle formula for cosine again to express cos2 2x.

Z

1

sin x dx =

4

Z

=

4

=

12.

Z

cos 4x = 2 cos2 2x

1

Z

2

1

2

=)

cos2 2x =

sin 2x +

1

8

1

4

1

2 cos 2x + cos2 2x dx

1

(cos 4x + 1)

2

Z

1

1 2 cos 2x + cos 2x dx =

1

4

Z

1 1

1

1

cos 2x + cos 4x +

dx =

4 2

8

8

1

2

Z

3

sin 4x + x + C =

8

1

(cos 4x + 1)

2

1

1

3

cos 2x + cos 4x +

2

8

8

2 cos 2x +

dx

dx

1

1

3

sin 2x +

sin 4x + x + C

4

32

8

sin5 x dx

Solution: This method works with odd powers of sin x or cos x. We will separate one factor of sin x from the

rest which will be expressed in terms of cos x.

Z

Z

Z

Z

Z

2

2

5

4

4

2

sin x dx =

sin x sin x dx = sin x sin x dx = sin x sin x dx = sin x 1 cos2 x dx

Z

=

sin x 1 2 cos2 x + cos4 x dx

We proceed with substitution. Let u = cos x: Then du = sin xdx.

Z

Z

Z

sin5 x dx =

sin x 1 2 cos2 x + cos4 x dx =

1 2 cos2 x + cos4 x (sin xdx)

Z

Z

2

1 5

=

1 2u2 + u4 ( du) =

1 + 2u2 u4 du = u + u3

u +C

3

5

1

2

=

cos x + cos3 x

cos5 x + C

3

5

13.

Z

a2

1

dx

+ b2 x2

Z

1

dx = tan

+1

a2 x2 = b2 u2 . This would be convenient because then

Solution:

The basic integral here is

x2

1x

+ C.

We need a substitution under which

1

1

1

1

= 2 2

= 2

2

2

2

+b

b u +b

b u +1

a2 x2

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

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