Download 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.
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