solutions Brief for (parts of) Homeworks 3 and 4
Homework 4
/6
/6
3
0
0
= u
u3
3
2. I =
1/2
=
0
(1 u2 ) du
(1 sin x) cos x dx =
cos x dx =
1.
sin /6
2
sin 0
1
11
1
=.
2 24
24
sec3 x dx =
sec x sec2 x dx. Integrate by parts with u = sec x,
dv = sec2 x dx, so du = sec x tan x dx and v = tan x. Then
I = sec x tan x
sec x tan2 x dx
I = sec x tan x
sec x(sec2 x 1) dx
I = sec x tan x
sec3 x dx +
sec x dx
I = sec x tan x I + ln | sec x + tan x|
2I = sec x tan x + ln | sec x + tan x|
1
I=
sec x tan x + ln | sec x + tan x| + C
2
3.
=
x3
1000 tan3
dx =
10 sec2 d
2
x2 + 100
100 tan + 100
3
1000 tan
10 sec2 d = 1000 tan3 sec d
10 sec
= 1000
tan2 sec tan d = 1000
= 1000
(u2 1) du = 1000
(sec2 1) sec tan d
u3
sec3
u = 1000
sec .
3
3
1
x2 + 100
= sec . The nal answer is then
10
1 (x2 + 100)3/2
x2 + 100
1000
+C
3
1000
10
x
Since
= tan , it follows that
10
which can probably be written in other ways as well.
4. We look for A and B that satisfy
x2
x9
A
B
x9
=
=
+
+ 3x 10
(x 2)(x + 5)
x2 x+5
which implies x 9 = A(x + 5) + B (x 2). Either expanding and comparing,
or plugging in special values like x = 2 and x = 5, we nd that A = 1
and B = 2. Therefore the integral is
2
1
dx = 2 ln |x + 5| ln |x 2| + C.
x+5 x2
Homework 3
5
xex dx, try parts with u = x, dv = ex dx. Then du = 1 dx and
1. For
0
v = ex . Then the integral is
xex
5
5
0
5
ex dx = 5e5 +
ex dx = 5e5 + ex
0
= 5e5 + ex
0
0
0
5
= 5e5 + 1 e5 = 1 6e5 .
5
2. For
t sin 2t dt, try parts with u = t, = dv sin 2t dt. Then du = dt and
v = (1/2) cos 2t. Then the integral is
1
t
1
t
cos 2t dt = cos 2t +
cos 2t
2
2
2
2
t
1
= cos 2t + sin 2t + C .
2
4
2
cos 2t dt
3. For
ln x dx, try parts with u = ln x, dv = 1 dx. Then du = (1/x) dx
and v = x. Then the integral is
1
x ln x x dx = x ln x 1 dx = x ln x x + C.
x
4. For
arcsin x dx, try parts with u = arcsin x, dv = 1 dx. Then
1
dx and v = x. Then the integral is
1 x2
x
1
2x
x arcsin x
dx = x arcsin x +
dx
2
1 x2
1 x2
1
1
du = x arcsin x + u = x arcsin x + 1 x2 + C .
= x arcsin x +
2
u
du =
ex cos 2x dx, try parts. We can start in either of two ways!
5. For I =
Say we choose u = ex and dv = cos 2x dx. Then du = ex dx and
v = (1/2) sin 2x. Then we have
1
1
1
1
ex sin 2x dx = ex sin 2x +
ex sin 2x dx.
I = ex sin 2x
2
2
2
2
Integrate by parts again, using a similar choice as before. Choose u = ex
and dv = sin 2x dx. Then du = ex dx and v = (1/2) cos 2x. Then
1
1
I = ex sin 2x +
2
2
1
ex cos 2x
2
1
+ ex cos 2x dx
2
1
1
1
I = ex sin 2x ex cos 2x
ex cos 2x dx
2
4
4
1
1
5
I = ex sin 2x ex cos 2x
4
2
4
2 x
1 x
I = e sin 2x e cos 2x + C .
5
5
/2
1
cos x dx =
2
/2
2
6.
0
(1 + cos 2x) dx =
0
3
1
sin 2x
x+
2
2
/2
=
0
.
4
sin6 x cos3 x dx =
7.
=
(u6 u8 ) du =
/2
cos5 x dx =
0
/2
(cos2 x)2 cos x dx =
0
0
(1 u2 )2 du =
x=0
=1
(1 sin2 x)2 cos x dx
1
x=/2
=
u6 (1 u2 ) du
sin7 x sin9 x
u7 u9
=
+ C.
7
9
7
9
/2
8.
sin6 x(1 sin2 x) cos x dx =
(1 2u2 + u4 ) du = u
0
21
15 10
3
8
+=
+
=.
35
15 15 15
15
4
2u3 u5
+
3
5
1
0