Practice Problems: Trig Integrals (Solutions) Written by Victoria Kala [email protected] November 9, 2014 The following are solutions to the Trig Integrals practice problems posted on November 9. R 1. sec xdx Note: This is an integral you should just memorize so you don’t need to repeat this process again. Solution:

Z

Z sec xdx =

sec x

sec x + tan x dx = sec x + tan x

Z

sec2 x + sec x tan x dx sec x + tan x

Let w = sec x + tan x, so du = (sec x tan x + sec2 x)dx: Z Z sec2 x + sec x tan x 1 dx = dw = ln |w| + C sec x + tan x w Plug back in w: Z sec xdx = ln | sec x + tan x| + C  2.

R

sec3 xdx

Solution: Rewrite:

Z

Z

3

sec 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: Z Z Z sec3 xdx = sec x tan x − sec x tan2 xdx = sec x tan x − sec x(sec2 x − 1)dx Z = sec x tan x −

sec3 xdx +

Z

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

1

3.

R

cos4 xdx

Solution: Since we have an even power of cos, we need to use the half angle identity: 2 Z Z Z  Z 1 1 4 2 2 cos xdx = (cos x) dx = (1 + cos 2x) dx = (1 + 2 cos 2x + cos2 2x)dx 2 4 Use half angle again:   Z  Z  1 1 1 3 1 1 + 2 cos 2x + (1 + cos 4x) dx = + 2 cos 2x + cos 4x dx 4 2 4 2 2   1 1 3 x + sin 2x + sin 4x + C = 4 2 8  4.

R

t sin2 tdt

Solution: Use half angle identity:  Z  Z Z  Z 1 1 2 t sin tdt = t (1 − cos 2t) dt = tdt − t cos 2tdt 2 2 The first integral is straightforward, use integration by parts (tabular method) on the second with u = t, dv = cos 2tdt:   Z 1 1 2 1 1 2 t sin tdt = t − t sin 2t − cos 2t + C 2 2 2 4  5.

R

3 sin√

√ x

x

dx √

1 1 x, so dw = √ dx ⇒ 2dw = √ dx: 2 x x Z Z Z Z 3√ sin x √ dx = 2 sin3 wdw = 2 sin w · sin2 wdw = 2 sin w(1 − cos2 w)dw x

Solution: Let w =

Let y = cos w, so dy = − sin wdw:   Z Z 1 3 2 2 2 sin w(1 − cos w)dw = −2 (1 − y )dy = −2 y − y 3 Plug back in w: 

   1 3 1 3 −2 y − y = −2 cos w − cos w 3 3 Plug back in x and add C:     √ √ 1 1 −2 cos w − cos3 w = −2 cos x − cos3 x + C 3 3 

2

6.

Rπ

sin2 t cos4 tdt

0

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.   Z π Z π Z π 1 2 2 4 2 2 2 (1 + cos 2t) dt sin t cos tdt = sin t cos t cos tdt = (sin t cos t) 2 0 0 0 1 2

π

Z



0

2 Z 1 1 π (sin 2t)2 (1 + cos 2t)dt sin 2t (1 + cos 2t)dt = 2 8 0 Z π  Z π 1 (sin 2t)2 dt + = (sin 2t)2 cos 2tdt 8 0 0

Let’s look at these integrals separately. The left integral we need to use half angle identity:   Z π Z 1 π 1 1 π π sin2 2tdt = t − sin 2t = (1 − cos 2t)dt = 2 0 2 2 2 0 0 Now let’s look at the right integral. Use the substitution w = sin 2t, then dw = 2 cos 2tdt: Z Z π 1 0 2 w dw = 0 (sin 2t)2 cos 2tdt = 2 0 0 So the final answer is:

Z

π

sin2 t cos4 tdt =

0

1 π π = 8 2 16

 7.

R π/2 0

(2 − sin θ)2 dθ

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

π/2 2

Z

(2 − sin θ) dθ = 0

π/2 2

(4 − 4 sin θ + sin θ)dθ = 0

Z

π/2 

= 0

Z 0

π/2



 1 1 4 − 4 sin θ + − cos 2θ dθ 2 2



π/2 9 1 9 1 9π − 4 sin θ − cos 2θ dθ = θ + 4 cos θ − sin 2θ = −4 2 2 2 4 4 0

 8.

R

cos2 x sin 2xdx

Solution: Z

cos2 x sin 2xdx =

Z

cos2 x · 2 sin x cos xdx = 2

3

Z

sin x cos3 xdx

Let w = cos x, dw = − sin xdx: Z Z 1 1 2 sin x cos3 xdx = −2 w3 dw = − w4 + c = − cos4 x + C 2 2  9.

R

tan x sec3 xdx

Solution:

Z

Z

3

tan x sec xdx =

sec2 x · sec x tan xdx

Let w = sec x, dw = sec x tan xdx: Z Z 1 1 2 sec x · sec x tan xdx = w2 dw = w3 + C = sec3 x + C 3 3  10.

R

x sec x tan xdx

Solution: Use integration by parts with u = x, dv = sec x tan xdx. Then du = dx, v = sec x: Z Z x sec x tan xdx = x sec x − sec xdx = x sec x − ln | sec x + tan x| + C  11.

R

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:

Z

Z csc xdx =

csc x

csc x − cot x dx = csc x − cot x

Z

csc2 x − csc x cot x dx csc x − cot x

Let w = csc x − cot x. Then dw = (− csc x cot x + csc2 x)dx: Z Z csc2 x − csc x cot x 1 dx = dw = ln |w| + C = ln | csc x − cot x| + C csc x − cot x w  12.

R

cot3 xdx

Solution: Z Z Z Z Z cot3 xdx = cot x cot2 xdx = cot x(csc2 x − 1)dx = cot x csc2 xdx − cot xdx

4

Z =

Z 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: Z Z 1 1 csc x · csc x cot xdx = − wdw = − w2 = − csc2 x 2 2 Now let’s look at the second integral. Rewrite it and let y = sin x so dy = cos xdx: Z Z Z 1 cos x dx = dy = ln |y| = ln | sin x| cot xdx = sin x y Now combine the two answers and add C: Z 1 cot3 xdx = − csc2 x + ln | sin x| + C 2  13.

R

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. Z

1 sin 8x cos 5xdx = 2

Z

   1 1 1 (sin 3x + sin 13x) dx = − cos 3x − cos 13x + C 2 3 13

 14.

R

cos πx cos 4πxdx

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 = 12 [cos(A − B) + cos(A + B)]. Z

1 cos πx cos 4πxdx = 2

Z

 Z  1 (cos (−3πx) + cos 5πx)dx = (cos 3πx + cos 5πx)dx 2   1 1 1 = sin 3πx + sin 5πx + C 2 3π 5π

 15.

R π/6 √ 0

1 + cos 2xdx

Solution: This is using the half identity backwards.

5

Z

π/6

√

Z

π/6

r

1 + cos 2xdx =

0

0

Z π/6 √ √ Z π/6 1 2 2 · (1 + cos 2x)dx = 2 cos xdx = 2 cos xdx 2 0 0 √ π/6 √ 2 = = 2 sin x 2 0

 16.

R π/4 √ 0

1 − cos 4θdθ

Solution: This is using the half identity backwards. Z π/4 Z π/4 r Z π/4 p √ Z π/4 √ 1 1 − cos 4θdθ = 2 · (1 − cos 4θ)dθ = 2 sin2 2θdθ = 2 sin 2θdθ 2 0 0 0 0 √   √ 1 2 π/4 = = 2 − cos 2θ 2 2 0  17.

R

1−tan2 x sec2 x dx

Solution: Z

1 − tan2 x dx = sec2 x

Z

(cos2 x − sin2 x)dx =

Z cos 2xdx =

1 sin 2x + C 2

 18.

R

dx cos x−1

Solution: Multiply by the conjugate: Z Z Z Z 1 cos x + 1 cos x + 1 cos x + 1 dx = · dx = dx = dx cos x − 1 cos x − 1 cos x + 1 cos2 x − 1 − sin2 x Z = (− csc x cot x − csc2 x)dx = csc x + cot x + C  19.

R

x tan2 xdx

Solution: Use the identity tan2 x = sec2 x − 1: Z Z Z Z 2 2 2 x tan xdx = x(sec x − 1)dx = x sec xdx − xdx

6

The last integral is no problemo. The first integral we need to use integration by parts. Let u = x, dv = sec2 x. Then du = dx, v = tan x, so: Z Z x sec2 xdx = x tan x − tan xdx You can rewrite the last integral as − ln | cos x|, so: Z

R

sin x cos x dx

and use the substitution w = cos x.

R

tan xdx =

x sec2 xdx = x tan x + ln | cos x|

Plug that into the original integral: Z 1 x tan2 xdx = x tan x + ln | cos x| − x2 + C 2  20.

R

x sin2 (x2 )dx

Solution: Let w = x2 , dw = 2xdx:   Z Z Z 1 1 1 1 1 2 2 2 sin wdw = · (1 − cos 2w)dw = w − sin 2w + C x sin (x )dx = 2 2 2 4 2 =


1 1 2 (w − sin w cos w) + C = x − sin (x2 ) cos (x2 ) + C 4 4

 21. Find the area of the region bounded by the given curves: y = sin3 x, y = cos3 x, π/4 ≤ x ≤ 5π/4 Solution: To find the area we need to subtract the bottom function from the top function and then integrate over our domain: Z 5π/4 Z 5π/4 Z 5π/4 A= (sin3 x − cos3 x)dx = sin3 xdx − cos3 xdx π/4

π/4

Let’s look at the first integral: Z 5π/4 Z sin3 xdx = π/4

5π/4

sin x · sin2 xdx =

π/4

π/4

Z

5π/4

sin x(1 − cos2 x)dx

π/4

Let w = cos x, dw = − sin xdx: Z

5π/4

sin x(1 − cos2 x)dx = −

π/4

 =

Z

√ − 2/2

√

(1 − w2 )dw = 2/2

√  √ √ 1 3 2/2 2 w− w √ = 2− 3 6 − 2/2 7

Z

√ 2/2

(1 √ − 2/2

− w2 )dw

Now let’s look at the second integral: Z

5π/4

cos3 xdx =

π/4

Z

5π/4

cos x · cos2 xdx =

Z

π/4

5π/4

cos x(1 − sin2 x)dx

π/4

Let y = sin x, dy = cos xdx: Z

5π/4

Z

2

cos x(1 − sin x)dx =

√ − 2/2

(1 − y 2 )dy

√ 2/2

π/4

√   √ √ 1 3 − 2/2 2 =− 2+ = y − y √ 3 6 2/2 Now plug this back in: Z

5π/4

sin3 xdx −

A= π/4

Z

5π/4

cos3 xdx =

√

π/4

√ ! 2 2− − 6

√ − 2+

√ ! √ √ √ 2 2 5 2 =2 2− = 6 3 3

 22. Find the volume obtained by rotating the region bounded by the given curves about the specified axis: y = sec x, y = cos x, 0 ≤ x ≤ π/3 about y = −1. Solution: Use the washer method. The outer area is given by A = πr2 = π(sec x + 1)2 , and the inside area is given by A = πr2 = π(cos x + 1)2 : π/3

Z

Z

π/3

((sec x + 1)2 − (cos x + 1)2 )dx

(outer area − inside area)dx = π

V = 0

0

π/3

π/3

 1 1 =π (sec x+2 sec x−cos x−2 cos x)dx = π sec x + 2 sec x − − cos 2x − 2 cos x dx 2 2 0 0   1 1 π/3 = π tan x + 2 ln | sec x + tan x| − x − sin 2x − 2 sin x 2 4 0 √ !   √ π π 3 π π π 1 2π π = π tan + 2 ln sec + tan − − sin − 2 sin = π 2 ln(2 + 3) − − 3 3 3 6 4 3 3 6 8 Z

2

Z

2



8



2