Sample Problems - mathcs.wilkes.edu

[Pages:5]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

Lecture Notes

Trigonometric Integrals 1

page 6

So we will pursue this substitution.

We

solve

a2x2

=

b2u2

for

a

possible

value

of

u

and

obtain

u

=

a x.

Then

b

a

b

du = dx and so du = dx.

b

a

Z

Z

1

1

a2x2 + b2 dx =

b2u2 + b2

Z

Z

b du =

a

1 1b

b

b2 u2 + 1 a du = ab2

1 u2 + 1

1 du = tan

ab

1u+C

= 1 tan 1 b x + C

ab

a

Z 14. p 1 dx

a2 x2 Z

Solution: The basic integral here is p 1 dx = sin 1 x + C. 1 x2

x2 = a2u2. This would be useful because then

We need a substitution under which

p 1 =p 1

=p 1

= p1

a2 x2

a2 a2u2

a2 (1 u2) a 1 u2

So we will pursue this substitution. We solve x2 = a2u2 for a possible value of u and obtain x = au and

dx = adu.

Z

Z

Z

Z

p 1 dx = p 1

(adu) = p a du = p 1 du = sin 1 u+C = sin 1 x + C

a2 x2

a2 a2u2

a 1 u2

1 u2

a

Z

p

15. secp( x) dx

x

p Let u = x.

Then du =

p1 dx.

2x

Z

p

secp( x) dx

=

Z 2

p

Z

secp( x) dx = 2

sec

p x

x

2x

p

p

= 2 ln jsec ( x) + tan ( x)j + C

Z

p1 dx = 2 2x

sec u du = 2 ln jsec u + tan uj + C

Z=3 p

16.

1 + cos 2x dx

0

Solution: We will yet again use the double angle formula for cosine, this time to eliminate the square root.

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

Z=3 p

Z=3 p

p Z=3 p

p Z=3

1 + cos 2x dx =

2 cos2 x dx = 2

cos2 x dx = 2 jcos xj dx

0

0

hi

0

0

Since f (x) = cos x is positive on 0; 3 ; we can simplify jcos xj = cos x

p Z=3

p Z=3

p

2 jcos xj dx = 2 cos x dx = 2

0

0

!

=3

p

sin x

= 2 sin

0

3

p

p 3

sin 0 = 2 2

!p 6

0=2

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

Lecture Notes

Trigonometric Integrals 1

page 7

Z=2 p

17.

1 cos x dx

0

Solution:

We substitute

cos 2 = 1 2 sin2 =) = x into this and obtain

2

2 sin2 x = 1 2

2 sin2 = 1 cos x

cos 2

Z=2 p 1

cos x

dx =

Z=2 r 2 sin2

x

p Z=2 dx = 2

x sin

dx

2

2

0

x

h i0

0

x

x

Since f (x) = sin is non-negative on 0; ; we can simplify sin = sin

2

2

2

2

p Z=2 x

p

!

x =2

p

!

x =2

p

2 sin dx = 2 2 cos

= 2 2 cos

= 2 2 cos

2

20

20

4

0

p

p 2

!

p

p

= 22

1 = 2+2 2= 2 2 2

2

Z 18. tan3 x dx

cos 0

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

Z tan3 x dx

= =

Z Z

sin3 x

Z

sin2 x

Z 1

cos2 x

Z 1

u2

cos3 u2 u3

x dx = 1 u3 du =

sin Z

x

cos3

1

u

u

x

3

dx =

cos3 x

u2

du = ln juj

2

sin xdx = + C = ln juj

+

u3 1 2u2 +

( C

Z u2 1

du) =

u3 du

= ln jcos xj + 1 sec2 x + C 2

Z

19. sin 7x cos 3x dx

Solution: We wil use the product-to-sum identities to trasform this product into a sum. We write the sine formula for the sum and the di?erence of these two angles.

sin 10x = sin (7x + 3x) = sin 7x cos 3x + cos 7x sin 3x sin 4x = sin (7x 3x) = sin 7x cos 3x cos 7x sin 3x

We will add the two equations

sin 10x + sin 4x = 2 sin 7x cos 3x 1 2 (sin 10x + sin 4x) = sin 7x cos 3x

1

We can very easily integrate (sin 10x + sin 4x)

Z

Z2 1

1Z

sin 7x cos 3x dx =

(sin 10x + sin 4x) dx = sin 10x + sin 4x dx

2

2

= 1 1 ( cos 10x) + 1 1 ( cos 4x) + C = 1 cos 10x

2 10

24

20

1 cos 4x + C 8

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

Lecture Notes

Trigonometric Integrals 1

page 8

Z 20. sin 10x sin 4x dx

Solution: We wil use the product-to-sum identities to trasform this product into a sum. We write the cosine formula for the sum and the di?erence of these two angles.

cos 14x = cos (10x + 4x) = cos 10x cos 4x sin 10x sin 4x cos 6x = cos (10x 4x) = cos 10x cos 4x + sin 10x sin 4x

We will subtract the ...rst equation from the ...rst one

cos 6x cos 14x = 2 sin 10x sin 4x 1

(cos 6x cos 14x) = sin 10x sin 4x 2

We can very easily integrate 1 (cos 6x cos 14x) 2

Z

Z

Z

sin 10x sin 4x dx = 1 (cos 6x cos 14x) dx = 1 cos 6x

2

2

cos 14x dx

= 1 1 (sin 6x) 1 1 (sin 14x) + C = 1 sin 6x 1 sin 14x + C

26

2 14

12

28

For more documents like this, visit our page at and click on Lecture Notes. E-mail questions or comments to mhidegkuti@ccc.edu.

c copyright Hidegkuti, Powell, 2012

Last revised: December 8, 2013

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

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

Google Online Preview   Download