Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 6.2
Exercise 3
Let u = cos x then du = − sin xdx. Thus,
Z π/2
Z π/2
7
5
(1 − cos2 x)3 sin x cos5 xdx
sin x cos xdx =
0
0
Z 1
Z 1
2 3 5
=
(1 − u ) u du =
(1 − 3u2 + 3u4 − u6 )u5 du
0
0
u6
3
3
1
=
− u8 + u10 − u12
6
8
10
12
1
=
120
1
0
Exercise 5
Using the trigonometric identity
1 + cos 2x
2
cos2 x =
we find
Z
π/2
2
π/2
Z
cos xdx =
0
0
1
=
2
Z
1 + cos 2x
dx
2
π/2
(1 + cos 2x)dx
0
1
sin 2x π/2
= x+
2
2
0
π
=
4
Exercise 7
Letting u = 2x and using the trigonometric identity
cos2 u =
1 + cos 2u
2
1
we find
Z
0
π
Z
Z
1 2π
1 2π 1 + cos 2u 2
2 2
cos 2xdx =
[cos u] du =
du
2 0
2 0
2
Z
1 2π
=
(1 + 2 cos 2u + cos2 2u)du
8 0
u sin 4u 2π
1
= u + sin 2u + +
8
2
8
0
3π
=
8
4
Exercise 11
Using integration by parts with u = t and v 0 = sin2 t, we have
Z Z
t
sin 2t
1
sin 2t
2
t sin tdt =
t−
−
t−
dt
2
2
2
2
t
sin 2t
t2 cos 2t
=
t−
− −
+C
2
2
4
8
t
cos 2t
t2
+C
= − sin 2t −
4
4
8
Exercise 17
Letting u = sec x so that du = tan x sec xdx, we have
Z
Z
u3
tan x sec3 xdx = u2 du =
+C
3
sec3 x
=
+C
3
Exercise 21
Letting u = tan x so that du = sec x2 dx, we have
Z
Z
4
6
tan x sec xdx = u4 (u2 + 1)2 du
Z
u9 2 7 u5
= (u8 + 2u6 + u4 )du =
+ u +
+C
9
7
5
tan9 x 2
tan5 x
=
+ tan7 x +
+C
9
7
5
2
Exercise 27
Letting u = tan x so that du = sec2 xdx and using Example 7 of the book,
we have
Z
Z
Z
5
3
2
tan xdx = tan x tan xdx = tan3 x(sec2 x − 1)dx
Z
Z
3
2
= tan x sec xdx − tan3 xdx
=
tan4 x tan2 x
−
+ ln | sec x| + C
4
2
Exercise 33
Using the fact that
(csc x − cot x)0 = − csc x cot x + csc2 x = csc x(csc x − cot x)
we have
Z
csc x(csc x − cot x)
dx
csc x − cot x
Z
(csc x − cot x)0
=
dx
csc x − cot x
= ln | csc x − cot x| + C
Z
csc xdx =
Exercise 39
√
Let x = 2 sin θ with − π2 < θ < π2 . Then 4 − x2 = 2 cos θ. We have
Z
Z
2 cos θ
dx
√
=
dθ
2
2
4 sin2 θ(2 cos θ)
x 4−x
Z
1
1
=
csc2 θdθ = − cot θ
4
4
√
2
4−x
=−
+C
4x
3
Exercise 40
√
Let x = 2 tan θ with − π2 < θ < π2 . Then x2 + 4 = 2 sec θ. We have
Z
Z
8 tan3 θ(2 sec2 θ)
x3
√
dx =
dθ
2 sec θ
x2 + 4
Z
Z
tan3 θ sec θdθ = 8 (sec2 θ − 1) tan θ sec θdθ
3
Z
sec θ
2
=8 (u − 1)du = 8
− sec θ + C
3
p
p
1
= (x2 + 4) x2 + 4 − 4 x2 + 4 + C
3
=8
Exercise 41
√
Let x = 2 sec θ with 0 < θ < π2 or π < θ < 3π
x2 − 4 = 2 tan θ. We
2 . Then
have
Z
Z √ 2
x −4
2 tan θ(2 sec θ tan θ)
dx =
dθ
x
2 sec θ
Z
Z
=2 tan2 θdθ = 2 (sec2 θ − 1)dθ
x
p
=2 tan θ − 2θ + C = x2 − 4 − 2 sec−1
+C
2
4
Exercise 45
Let x = a tan θ with − π2 < θ <
a sec2 θdθ. Hence,
Z
0
a
π
2.
Then
√
a2 + x2 = a sec θ and dx =
π/4
a sec2 θ
dθ
a3 sec3 θ
0
(a2 + x2 )
√
1
2
π/4
= 2 sin θ]0 = 2
a
2a
Z
dx
3
2
=
Exercise 49
√
Let 2x = sin θ with − π2 < θ < π2 . Then 1 − 4x2 = cos θ and 2dx = cos θdθ.
Thus,
Z
Z p
1
2
1 − 4x dx =
cos2 θdθ
2
Z
1
θ sin 2θ
=
(1 + cos 2θ)dθ = +
+C
4
4
8
θ 2 sin θ cos θ
= +
+C
4
8
√
x 1 − 4x2
1
= sin−1 (2x) +
+C
4
2
Exercise 51
√
Let x = 3 sec θ so that x2 − 9 = 3 tan θ and dx = 3 sec θ tan θdθ. Thus,
Z √ 2
Z
x −9
1
tan2 θ
dx
=
dθ
x3
3
sec2 θ
Z
1
θ
1
=
sin2 θdθ = −
sin 2θ + C
3
6 12
√x2 − 9
1
−1 x
= sec
−
+C
6
3
2x
5
Exercise 53
√
Let 5x = 3 sin θ with − π2 < θ < π2 . Then 9 − 25x2 = 3 cos θ and 5dx =
3 cos θdθ. Thus,
Z 0.6
Z π/2
x2
9
√
dx =
sin2 θdθ
125 0
9 − 25x2
0
9 θ sin 2θ π/2
−
=
125 2
4
0
9 π
9π
=
[ ]=
125 4
500
Exercise 57
√
Let x = tan θ so that x2 + 1 = sec θ and dx = sec2 θdθ. Thus,
Z √ 2
Z
sec3 θ
x +1
dx =
dθ
x
tan θ
Z tan2 θ + 1
=
sec θdθ
tan θ
Z 1
sec θdθ
=
tan θ +
tan θ
Z
Z
sin θ
=
dθ
+
csc θdθ
cos2 θ
= sec θ + ln | csc θ − cot θ| + C
√
x2 + 1 1 p
= x2 + 1 + ln − +C
x
x
Exercise 65
We have
Z
f (t) =
t
sin ωs cos2 ωsds = −
0
t
1
cos3 ωs
3ω
0
1
=
1 − cos3 ωt
3ω
Exercise 66
(a) Recall that the average of f (x) in [a, b] is given by
Z b
1
f (x)dx.
b−a a
6
Now, one cycle takes
1
60
seconds. We are asked to find
"
We have
Z
Z 1/60
E(t)2 dt =24, 025
0
1
1
60 − 0
Z
#1
1/60
2
2
E(t) dt
.
0
1/60
24, 025
1
sin (120πt)dt =
60πt − sin (240πt)
120π
4
0
0
24, 025
4805
=
(π) =
.
120π
24
1/60
2
Thus,
"
RM S =
1 4805
1
24
60 − 0
#1
2
≈ 109.60.
(b) We must solve the equation
2202 =
1
1
60 − 0
Z
1/60
A2 sin2 (120πt)dt.
0
That is,
2202 =
60A2
120
which yields A ≈ 311.13
Exercise 67
Using the substitution x = sec θ, we have
Z 7√ 2
Z −1
1
x −1
1 sec 7
dx =
tan2 θdθ
7−1 1
x
6 sec−1 1
Z −1
1 sec 7
=
(sec2 θ − 1)dθ
6 sec−1 1
1
sec−1 7
= [tan θ − θ]sec
−1 1
6
i7
1 hp 2
x − 1 − sec−1 x
=
6
1
1 √
−1
= ( 48 − sec 7)
6
7
Exercise 70
From Exercise 45, we find
Z
dx
(x2
+
3
b2 ) 2
=
x
sin θ
+C = √
+ C.
2
2
b
b x2 + b2
Thus,
"
#
L−a
λb
x
λ
L−a
a
√
p
E(P ) =
=
+√
4π0 b2 x2 + b2 −a
4π0 b
a2 + b2
(L − a)2 + b2
8