Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
Solutions to Assignment 6.1
Exercise 3
We have
u(x) = x v 0 (x) = cos 5x
u0 (x) = 1 v(x) = sin55x .
Using the method of integration by parts, we find
Z
Z
x
1
x cos 5xdx = sin 5x −
sin 5xdx
5
5
x
1
= sin 5x +
cos 5x + C
5
25
Exercise 5
We have
u(t) = t
v 0 (t) = e−3t
−3t
0
u (t) = 1 v(t) = − e 3 .
Using the method of integration by parts, we find
Z
Z
t −3t 1
−3t
te
=− e
+
e−3t dt
3
3
t
1
= − e−3t − e−3t + C
3
9
Exercise 9
We have
u(x) = ln (2x + 1) v 0 (x) = 1
2
u0 (x) = 2x+1
v(x) = x.
Using the method of integration by parts, we
Z
Z
ln (2x + 1)dx =x ln (2x + 1) −
Z
=x ln (2x + 1) −
Z
=x ln (2x + 1) −
find
2x
dx
2x + 1
2x + 1 − 1
dx
2x + 1
Z
dx
dx +
2x + 1
ln (2x + 1)
=x ln (2x + 1) − x +
+C
2
1
Exercise 13
We have
u(θ) = e2θ
v 0 (θ) = sin 3θ
0
2θ
u (θ) = 2e
v(θ) = − cos33θ .
Using the method of integration by parts twice, we find
Z
Z
2
e2θ
2θ
cos 3θ +
e2θ cos 3θdθ
e sin 3θdθ = −
3
3
Z
e2θ
2 e2θ
2
2θ
=−
cos 3θ +
sin 3θ −
e sin 3θdθ
3
3 3
3
Z
2
e2θ
13
e2θ sin 3θdθ = −
cos 3θ + e2θ sin 3θ + C
9
3
9
Z
2
3
e2θ sin 3θdθ = − e2θ cos 3θ + e2θ sin 3θ + C 0
13
13
Exercise 15
We have
1
u(x) = xe2x
v 0 (x) = (1+2x)
2
1
0
2x
u (x) = e (1 + 2x) v(θ) = − 2 (1 + 2x)−1 .
Using the method of integration by parts, we find
Z
Z
xe2x
xe2x
1
−1
dx
=
−
(1
+
2x)
+
e2x dx
(1 + 2x)2
2
2
xe2x
e2x
=−
+
+C
2(1 + 2x)
4
e2x
+C
=
4(1 + 2x)
Exercise 19
We have
u(r) = ln r v 0 (r) = r3
4
v(r) = r4 .
u0 (r) = 1r
Using the method of integration by parts, we find
Z
1
3
3
Z
1 3 3
r4
r ln rdr =
ln r −
r dr
4
4 1
1
4
3
r
r4
=
ln r −
4
16 1
81
= ln 3 − 5
4
3
2
Exercise 23
We have
u(x) = cos−1 x v 0 (x) = 1
1
u0 (x) = − √1−x
v(x) = x.
2
Using the method of integration by parts, we find
1
2
Z
cos
−1
Z
1
2
x
dx
1 − x2
0
h
i1
p
2
−1
2
= x cos x − 1 − x
0
√
π
3
= −
+1
6
2
xdx = x cos
0
−1
1
x02 +
√
Exercise 25
We have
u(x) = (ln x)2 v 0 (x) = 1
x
u0 (x) = 2 ln
v(x) = x.
x
Using the method of integration by parts, we find
Z 2
Z 2
2
2 2
(ln x) dx = x(ln x) 1 − 2
ln xdx
1
1
2
= x(ln x)2 − 2x ln x + 2x 1
=2(ln 2)2 − 4 ln 2 + 2
Exercise 29
Letting x = θ2 , the given integral reduces to
√
Z
π
1
√ θ cos θ dθ = 2
π/2
3
2
Z
π
x cos xdx.
π
2
We have
u(x) = x v 0 (x) = cos x
u0 (x) = 1 v(x) = sin x.
Using the method of integration by parts, we find
Z
Z
x cos xdx = x sin x − sin xdx = x sin x + cos x + C.
Thus,
√
Z
√
π
θ3 cos θ2 dθ =
π/2
1
1
π
[x sin x + cos x]ππ = − 1 +
2
2
2
2
3
Exercise 32
(a) We have
u(x) = cosn−1 x
v 0 (x) = cos x
u0 (x) = −(n − 1) cosn−2 x sin x v(x) = sin x.
Using the method of integration by parts, we find
Z
Z
n
n−1
cos xdx = cos
x sin x + (n − 1) cosn−2 x sin2 xdx
Z
Z
n−1
n−2
= cos
x sin x + (n − 1) cos
xdx − (n − 1) cosn xdx
Z
Z
n
n−1
n cos xdx = cos
x sin x + (n − 1) cosn−2 xdx
Z
Z
cosn−1 x sin x n − 1
n
cos xdx =
+
cosn−2 xdx.
n
n
(b) Letting n = 2 in (a), we find
Z
sin 2x x
+ + C.
cos2 xdx =
4
2
(c) Letting n = 4 in (a) and using (b), we find
Z
Z
cos3 x sin x 3
4
cos xdx =
+
cos2 xdx
4
4
cos3 x sin x 3 sin 2x x
=
+
+
+C
4
4
4
2
Exercise 41
The average of a function f (x) in the interval [a, b] is given by
We have
u(x) = x v 0 (x) = sec2 x
u0 (x) = 1 v(x) = tan x.
Using the method of integration by parts, we find
Z
Z
2
x sec xdx =x tan x − tan xdx
Z
sin x
=x tan x −
dx
cos x
=x tan x + ln | cos x| + C.
1
b−a
The final answer is
Z π
π
4 4
4
2
x sec2 xdx = [x tan x + ln | cos x|]04 = 1 − ln 2
π 0
π
π
4
Rb
a
f (x)dx.
Exercise 42
We are asked to find h(60) where h(t) is the height t seconds after liftoff.
2
Since h(0) = 0, by letting x = 1 − 375
t and using Example 2, we have
60 2
−9.8t − 3000 ln 1 −
t
dt
h(60) =
375
0
Z 17
25
2 60
= −4.9t 0 + 562, 500
ln xdx
Z
1
17
60
= −4.9t2 + 562, 500 [x ln x − x] 25
1
0
≈14, 844.10
Exercise 43
Assuming s(0) = 0, using integration by parts twice, we have
Z
s(t) =
t
2 −s
s e
ds =
0
t
−s2 e−s 0
Z
+2
t
se−s ds
0
t
= −s2 e−s − 2se−s − 2e−s 0
= − t2 e−t − 2te−t − 2e−t + 2
Exercise 45
Letting u(x) = x and v 0 (x) = f 00 (x), using integration by parts, we find
Z
1
4
Z 4
4
xf (x)dx = xf (x) 1 −
f 0 (x)dx
1
0
4
= xf (x) − f (x) 1
00
0
=4f 0 (4) − f (4) − f 0 (1) + f (1) = 2
5