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Lecture Notes

Trigonometric Integrals 1

page 1

Sample Problems

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

Z 1. sin x dx

Z 8. csc x dx

Z 15.

p

secp(

x) dx

x

Z 2. cos 5x dx

Z 3. cos x sin4 x dx

Z 9. sin2 x dx

Z 10. sin3 x dx

Z=3 p

16.

1 + cos 2x dx

0

Z 4. csc2 x dx

Z 11. sin4 x dx

Z=2 p

17.

1

0

cos x dx

Z 5. tan x dx

Z 12. sin5 x dx

Z 18. tan3 x dx

Z 6. cot x dx

Z 1

13. a2 + b2x2 dx

Z 19. sin 7x cos 3x dx

Z 7. sec x dx

Z 14. p 1 dx

a2 x2

Z 20. sin 10x sin 4x dx

Z 1. cos 3x dx

Z

2. sin 4x

dx

5

Z

3. sec tan d

Z 4. sec2 d

Z 5. x tan x2 dx

Z 6. cot (2x ) dx

Z 7. cos2 x dx

Practice Problems

Z 8. cos2 (2x) dx

Z 9. cos3 x dx

Z 10. cos4 x dx

Z 11. cos5 x dx

Z 12. sin x cos5 x dx

Z 13. sin3 x cos5 x dx

Z 14. tan2 x dx

c copyright Hidegkuti, Powell, 2012

Z=6 p

15.

1

0

cos 6x dx

Z 16. sin 2a cos 8a da

Z 17. cos b cos 11b db

Z 18. sin 6 sin 14 d

Z 19. cos 11m sin 3m dm

Last revised: December 8, 2013

Lecture Notes

Trigonometric Integrals 1

page 2

Sample Problems - Answers

1 1.) cos x + C 2.) 5 sin 5x + C

3.) 1 sin5 x + C 4.) cot x + C 5.) ln jcos xj + C = ln jsec xj + C 5

6.) ln jsin xj + C

7.) ln jsec x + tan xj + C

8.) ln jcsc x + cot xj + C

11 9.) 2 x 4 sin 2x + C

10.) 1 cos3 x cos x + C 3

13.)

1 tan 1

b x

+C

ab

a

11.) 1 sin 2x + 1 sin 4x + 3 x + C

4

32

8

12.) cos x + 2 cos3 x 1 cos5 x + C

3

5

14.) sin 1 x + C a

p

p

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

p 6

16.) 2

p 17.) 2 2 2

18.) 1 sec2 x + ln jcos xj + C

2

1

1

1

1

19.) cos 10x cos 4x + C 20.) sin 6x sin 14x + C

20

8

12

28

Practice Problems - Answers

1

1

1.) sin 3x + C 2.) cos 4x

+C

3

4

5

3.) sec + C

4.) tan + C

5.)

1 ln

sec

x2

+C

2

1 6.) ln jsin (2x )j + C

2

11 7.) x + sin 2x + C

24

11 8.) x + sin 4x + C

28

9.) sin x 1 sin3 x + C 3

31

1

10.) x + sin 2x + sin 4x + C

11.) 1 sin5 x 2 sin3 x + sin x + C

84

32

5

3

12.) 1 cos6 x + C 6

13.) 1 cos6 x + 1 cos8 x + C 14.) x + tan x + C

6

8

p 2

15.) 3

1

1

16.) cos 6a

cos 10a + C

12

20

1

1

17.) 20 sin 10b + 24 sin 12b + C

1

1

18.) 16 sin 8 40 sin 20 + C

1

1

19.) 16 cos 8m 28 cos 14m + C

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

Lecture Notes

Trigonometric Integrals 1

page 3

Sample Problems - Solutions

Z 1. sin x dx

Solution: This is a basic integral we know from di?ereZntiating basic trigonometric functions. Since

d

d

dx cos x = sin x, clearly dx ( cos x) = sin x and so sin x dx = cos x + C .

Z

2. cos 5x dx

d

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

du

and so 5 = dx.

Z

Z

du

1Z

1

cos 5x dx = cos u

= cos u du = sin 5x + C

55

5

Note: Once wZ e have enough practice, there is no need to perform this substitution in writing. simply write cos 5x dx = 1 sin 5x + C.

5 Z

3. cos x sin4 x dx

We can just

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

Z

Z

Z

cos x sin4 x dx = sin4 x (cos xdx) = u4 u = 1 u5 + C = 1 sin5 x + C

5

5

Z 4. csc2 x dx

Solution: We need to remember that d cot x = dx

Z

Z

csc2 x dx =

csc2 x. csc2 x dx =

cot x + C

Z 5. tan x dx

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

Z

Z

Z

Z

tan x dx = sin x dx = 1 (sin xdu) = 1 ( du) =

cos x

u

u

= ln (cos x) 1 + C = ln jsec xj + C

Z 1 du = u

ln juj + C =

ln jcos xj + C

Z 6. cot x dx

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

Z

Z

Z

Z

cot x dx =

cos x dx = sin x

1 (cos xdu) = u

1 u

du = ln juj + C =

ln jsin xj + C

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

Lecture Notes

Trigonometric Integrals 1

page 4

Z 7. sec x dx

Z

Z

sec x + tan x

Z sec2 x + sec x tan x

Solution: sec x dx = sec x

dx =

dx

sec x + tan x

sec x + tan x

From here we will use substitution.

Recall that

d sec x = sec x tan x and

d tan x = sec2 x.

dx

dx

sec x + tan x. Then du = sec x tan x + sec2 x dx.

Let u =

Z sec2 x + sec x tan x

Z 1

dx =

Z

sec2 x + sec x tan x dx =

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

sec x + tan x

u

u

Z 8. csc x dx

Z

Z

csc x + cot x

Z csc2 x + csc x cot x

Solution: csc x dx = csc x

dx =

dx

csc x + cot x

csc x + cot x

From here we will use substitution.

d Recall that csc x =

d csc x cot x and cot x =

csc2 x.

Let

dx

dx

u = csc x + cot x. Then du = csc2 x csc x cot x dx.

Z csc2 x + csc x cot x

Z 1

dx =

Z

csc2 x + csc x cot x dx =

1 ( du) =

Z 1 du =

ln juj + C

csc x + cot x

u

u

u

= ln jcsc x + cot xj + C

Z 9. sin2 x dx

Solution: Recall the double angle formula for cosine, cos 2x = 1 2 sin2 x. We solve this for sin2 x sin2 x = 1 (1 cos 2x) 2

Z

Z

Z

sin2 x dx =

1

1

(1 cos 2x) dx =

1 dx

2

2

11 = x sin 2x + C

24

Z 10. sin3 x dx

Z

1

cos 2x dx

= 2

x + C1

1 2 sin 2x + C2

Solution:

Z

Z

Z

sin3 x dx = sin x sin2 x dx = sin x 1 cos2 x dx

Let u = cos x: Then du = sin xdx

Z

Z

Z

Z

Z

sin3 x dx = sin x 1 cos2 x dx = 1 cos2 x (sin xdx) = 1 u2 ( du) = u2 1 du

= 1 u3 u + C = 1 cos3 x cos x + C

3

3

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

Lecture Notes

Trigonometric Integrals 1

page 5

Z 11. sin4 x dx

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

cos 2x = 1 2 sin2 x =) sin2 x = 1 (1 cos 2x) 2

Z

Z

Z

sin4 x dx = sin2 x 2 dx =

1 (1

2

Z 1

cos 2x) dx = (1

Z cos 2x)2 dx = 1

1

2 cos 2x + cos2 2x dx

2

4

4

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

cos 4x = 2 cos2 2x 1

=)

cos2 2x = 1 (cos 4x + 1) 2

Z

Z

Z

sin4 x dx = 1

1

2 cos 2x + cos2 2x

1 dx =

1

Z4

4Z

11

1

1

=

cos 2x + cos 4x + dx =

42

8

8

1 2 cos 2x + (cos 4x + 1) dx

2

1 cos 2x + 1 cos 4x + 3 dx

2

8

8

= 1 1 sin 2x + 1 1 sin 4x + 3 x + C = 1 sin 2x + 1 sin 4x + 3 x + C

22

84

8

4

32

8

Z 12. 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

sin5 x dx = sin x sin4 x dx = sin x sin4 x dx = sin x sin2 x 2 dx = sin x 1 cos2 x 2 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

=

1 2u2 + u4 ( du) =

1 + 2u2 u4 du = u + 2 u3 1 u5 + C 35

=

cos x + 2 cos3 x 1 cos5 x + C

3

5

Z 1

13. a2 + b2x2 dx

Z

Solution: The basic integral here is

1 x2 + 1

dx

=

tan

1 x + C.

a2x2 = b2u2. This would be convenient because then

We need a substitution under which

1

1

11

a2x2 + b2 = b2u2 + b2 = b2 u2 + 1

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

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