Math 152 Extra Homework 5 Answer
1.
3π
4
Z
5
3π
4
Z
3
sin x cos xdx =
π
2
π
2
Z
Let u = sin x, du = cos xdx =
√
2
2
sin5 x(1 − sin2 x) cos xdx
5
2
u (1 − u )du =
1
1 8 1 6
=
u − u
8
6
1 1
= − −
8 6
=−
Z
1
√
2
2
u7 − u5 du
1
1
8
√1
2
4
!
1
1 1 3
−
2
6 2
11
.
384
2.
Z
Z
Z
tan2 x · sec xdx = (sec2 x − 1) sec xdx = sec3 xdx − ln | sec x + tan x| + C
Z
Z
3
where sec xdx = sec2 x · sec xdx( Let u = sec x, dv = sec2 xdx)
Z
(du = sec x tan x, v = tan x) = sec x tan x − sec x tan2 dx
Z
= sec x tan x − sec x(sec2 x − 1)dx
Z
Z
= sec x tan x − sec3 xdx + sec xdx
Z
1
1
Thus
sec3 xdx = sec x · tan x + ln | sec x + tan x| + C
2
2
Z
1
1
Hence
tan2 x · sec xdx = sec x · tan x − ln | sec x + tan x| + C.
2
2
3.
Z
Z cos x + sin x
1
1
1
dx =
+
dx
2 sin x cos x
2
sin x cos x
Z
1
1
=
(csc x + sec x) dx = [ln | csc x − cot x| + ln | sec x + tan x|] + C
2
2
1
= ln |csc x · sec x − csc x + sec x − 1| + C
2
cos x + sin x
dx =
sin 2x
Z
2
4. Let x = 3 sec θ. Then dx = 3 sec θ tan θdθ. We obtain
Z
dx
√
=
2
x x2 − 9
Z
3 sec θ tan θdθ
√
9 sec2 θ 9 sec2 θ − 9
Z
Z
1
tan θ
1
=
dθ =
cos θdθ
9
sec θ · tan θ
9
1
= sin θ + C
9
√
x2 − 9
1
+ C.
= ·
9
x
5. Let x = 4 tan θ. Then dx = 4 sec2 θdθ and tan θ = x4 , sec θ =
Z
dx
√
=
2
x + 16
Z
√
x2 +16
.
4
We have the integral is
4 sec2 θdθ
p
16(1 + tan2 θ)
Z
sec θdθ = ln | sec θ + tan θ| + C
√
x2 + 16 x p
= ln + + C = ln | x2 + 16 + x| + C.
4
4
=
6. Let t = x2 . Then dt = 2xdx. We have
Z
x
Z p
Z
p
1
1 p
1 − x4 dx =
1 − t2 · dt =
1 − t2 dt.
2
2
Let t = sin θ. Then dt = cos θdθ, cos θ =
Z
√
1 − t2 .
Z p
Z
1 p
1 − t2 dt =
1 − sin2 θ · cos θdθ
2
Z
Z
1
1
2
=
cos θdθ =
(1 + cos 2θ) dθ
2
4
1
1
1
=
θ + sin 2θ = (θ + sin θ · cos θ) + C
4
2
4
p
1
=
arcsin t + t 1 − t2 + C
4
Z p
i
p
1h
x 1 − x4 dx =
arcsin(x2 ) + x2 1 − x4 + C
4
1
2
3
7. It is easy to get the area of the triangle P OQ = 12 r2 sin θ · cos θ.
√
Rr
Area of region RQR = r cos θ r − x2 dx. Let x = r cos α. Then dx = −r sin αdα. Thus
Z
r
Z
p
r − x2 dx =
r cos θ
0p
r2 (1 − cos2 α)(−r) sin αdα
θ
= −r2
Z
θ
0
sin2 αdα = −
r2
2
Z
0
[1 − cos(2α)]dα
θ
0
r2
1
α − sin(2α)
2
2
θ
1 2
1 2
= r θ − r sin θ · cos θ.
2
2
=−
Then the area of sector is
θr2
2 .
8. We sketch a graph of this cylinder. To find the percentage of the volume, we just need to
find the percentage of area the shaded region.
The left one is the original region, but if we rotate 90 degree clockwise. We will get right
√
one, which is easier to get the area by integration. We know y = 52 − x2 . Let x = 5 sin θ.
4
Then dx = 5 cos θdθ
Z 2p
Z
2
2
2
5 − x dx = 2
arcsin( 25 ) p
52 − (5 sin θ)2 · 5 cos θdθ
π
2
−5
Z
= 50
arcsin( 25 )
cos2 θdθ
− π2
Z
= 25
arcsin( 25 )
[1 + cos 2θ] dθ
− π2
arcsin( 2 )
= [25θ − 25 sin θ · cos θ]− π 5
2
√
π
2
= 25
+ 2 21
+ arcsin
2
5
√
25π
2
=
+ 2 21.
+ 25 arcsin
2
5
Then the percentage is
A
=
25π
25π
2
+ 25 arcsin
25π
2
5
√
+ 2 21
= 74.7684%.