Practice Problems: Trig Integrals (Solutions)

Practice Problems: Trig Integrals (Solutions)

Written by Victoria Kala vtkala@math.ucsb.edu November 9, 2014

The following are solutions to the Trig Integrals practice problems posted on November 9.

1. sec xdx

Note: This is an integral you should just memorize so you don't need to repeat this process again.

Solution:

sec x + tan x

sec2 x + sec x tan x

sec xdx = sec x

dx =

dx

sec x + tan x

sec x + tan x

Let w = sec x + tan x, so du = (sec x tan x + sec2 x)dx:

sec2 x + sec x tan x

1

dx = dw = ln |w| + C

sec x + tan x

w

Plug back in w:

sec xdx = ln | sec x + tan x| + C

2. sec3 xdx Solution: Rewrite:

sec3 xdx = sec x ? sec2 xdx

Use integration by parts. Let u = sec x, dv = sec2 xdx. Then du = sec x tan xdx and v = tan x:

sec3 xdx = sec x tan x - sec x tan2 xdx = sec x tan x - sec x(sec2 x - 1)dx

= sec x tan x - sec3 xdx + sec xdx = sec x tan x + ln | sec x + tan x| - sec3 xdx

Notice on the right side we have the same integral as what we started with, so move it over to the left side:

2 sec3 xdx = sec x tan x + ln | sec x + tan x|

Divide by 2 and add C: sec3 xdx = 1 (sec x tan x + ln | sec x + tan x|) + C 2

1

3. cos4 xdx

Solution: Since we have an even power of cos, we need to use the half angle identity:

cos4 xdx = (cos2 x)2dx =

1

2

1

(1 + cos 2x) dx =

(1 + 2 cos 2x + cos2 2x)dx

2

4

Use half angle again:

1

1

13

1

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

+ 2 cos 2x + cos 4x dx

4

2

42

2

13

1

=

x + sin 2x + sin 4x + C

42

8

4. t sin2 tdt

Solution: Use half angle identity:

t sin2 tdt =

1

1

t (1 - cos 2t) dt =

2

2

tdt - t cos 2tdt

The first integral is straightforward, use integration by parts (tabular method) on the second with u = t, dv = cos 2tdt:

t sin2 tdt = 1

1 t2

-

1 t

sin 2t

-

1

cos

2t

+C

22 2

4

5.

sin3

x

x

dx

Solution: Let w = x, so dw =

1 dx 2dw = 1 dx:

2x

x

sin3

x

dx = 2

sin3 wdw = 2

sin w ? sin2 wdw = 2

x

sin w(1 - cos2 w)dw

Let y = cos w, so dy = - sin wdw:

2 sin w(1 - cos2 w)dw = -2 (1 - y2)dy = -2 y - 1 y3 3

Plug back in w: -2

Plug back in x and add C:

y - 1y3 3

= -2

cos w - 1 cos3 w 3

-2

cos w - 1 cos3 w

= -2

cos

x

-

1

cos3

x

+C

3

3

2

6.

0

sin2

t

cos4

tdt

Solution: You can use half angle identity on this problem, but you would need to use it several

times. I don't think you would see a problem like this on your exam, but it is nice to practice

anyway.

sin2 t cos4 tdt =

sin2 t cos2 t cos2 tdt =

(sin t cos t)2

1 (1 + cos 2t)

dt

0

0

0

2

1

1

2

1

sin 2t (1 + cos 2t)dt =

(sin 2t)2(1 + cos 2t)dt

20 2

80

1 =

(sin 2t)2dt + (sin 2t)2 cos 2tdt

80

0

Let's look at these integrals separately. The left integral we need to use half angle identity:

sin2 2tdt = 1

1

1

(1 - cos 2t)dt = t - sin 2t

=

0

20

2

2

02

Now let's look at the right integral. Use the substitution w = sin 2t, then dw = 2 cos 2tdt:

So the final answer is:

(sin 2t)2 cos 2tdt

=

1

0

w2dw = 0

0

20

sin2 t cos4 tdt = 1

=

0

8 2 16

7.

/2 0

(2

-

sin

)2d

Solution: Multiply this all out and use half angle identity:

/2

(2 - sin )2d =

/2

(4 - 4 sin + sin2 )d =

/2

11

4 - 4 sin + - cos 2 d

0

0

0

22

/2 9

1

9

1

/2 9

=

- 4 sin - cos 2 d = + 4 cos - sin 2 = - 4

0

2

2

2

4

0

4

8. cos2 x sin 2xdx Solution:

cos2 x sin 2xdx = cos2 x ? 2 sin x cos xdx = 2 sin x cos3 xdx

3

Let w = cos x, dw = - sin xdx:

2

sin x cos3 xdx = -2

w3dw

=

1 -

w4

+

c

=

-

1

cos4

x

+

C

2

2

9. tan x sec3 xdx Solution:

tan x sec3 xdx = sec2 x ? sec x tan xdx

Let w = sec x, dw = sec x tan xdx: sec2 x ? sec x tan xdx =

w2dw = 1 w3 + C = 1 sec3 x + C

3

3

10. x sec x tan xdx Solution: Use integration by parts with u = x, dv = sec x tan xdx. Then du = dx, v = sec x:

x sec x tan xdx = x sec x - sec xdx = x sec x - ln | sec x + tan x| + C

11. csc xdx

Note: This is similar to the first problem. This is an integral you should just memorize so you don't need to repeat this process again.

Solution:

csc x - cot x

csc2 x - csc x cot x

csc xdx = csc x

dx =

dx

csc x - cot x

csc x - cot x

Let w = csc x - cot x. Then dw = (- csc x cot x + csc2 x)dx:

csc2 x - csc x cot x

1

dx = dw = ln |w| + C = ln | csc x - cot x| + C

csc x - cot x

w

12. cot3 xdx Solution: cot3 xdx = cot x cot2 xdx = cot x(csc2 x - 1)dx = cot x csc2 xdx - cot xdx

4

= csc x ? csc x cot xdx - cot xdx

Let's look at the first integral. Let w = csc x, then dw = - csc x cot xdx:

csc x ? csc x cot xdx = -

wdw

=

-

1 w2

=

1 -

csc2

x

2

2

Now let's look at the second integral. Rewrite it and let y = sin x so dy = cos xdx:

cos x

1

cot xdx =

dx = dy = ln |y| = ln | sin x|

sin x

y

Now combine the two answers and add C: cot3 xdx = - 1 csc2 x + ln | sin x| + C 2

13. sin 8x cos 5xdx

Solution: I don't think you would see a problem like this on your exam, but it is nice to

practice anyway. There is a trig identity listed on page 476 of your text: sin A cos B =

1 2

[sin

(A

-

B

)

-

sin

(A

+

B)].

You

can

also

derive

this

equation

yourself.

1 sin 8x cos 5xdx =

2

11

1

(sin 3x + sin 13x) dx = - cos 3x - cos 13x + C

23

13

14. cos x cos 4xdx

Solution: Similar to the previous problem, I don't think you would see a problem like this on

your

exam.

On

page

476

of

your

text

is

the

identity

cos A cos B

=

1 2

[cos(A

-

B)

+

cos(A

+

B)].

1 cos x cos 4xdx =

2

1 (cos (-3x) + cos 5x)dx =

2

(cos 3x + cos 5x)dx

11

1

=

sin 3x + sin 5x + C

2 3

5

15.

/6 0

1

+

cos

2xdx

Solution: This is using the half identity backwards.

5

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