intsubsolong_homework solution - Spring 2003 1. Solutions...

Spring 2003 Math 122 Integration by Substitution
1. Solutions for the integrals:
b. For the integral below, we make the substitution u = 2 x
2
- 3, so du = 4 x dx . It follows
that:
Z
x
p
2 x 2 - 3 dx =
1
4
Z
(
2 x
2
- 3
)
1
2
(4 x ) dx
=
1
4
Z
u
1
2
du
=
1
6
u
3
2
+ C
=
1
6
(2 x
2
- 3)
3
2
+ C
d. For the integral below, we make the substitution u = x
2
+ 4 x - 5, so du = (2 x + 4) dx .
It follows that:
Z
x + 2
( x 2 + 4 x - 5) 3
dx =
1
2
Z
(
x
2
+ 4 x - 5
)
- 3
(2 x + 4) dx
=
1
2
Z
u
- 3
du
= -
1
4
u
- 2
+ C
= -
1
4 ( x 2 + 4 x - 5)
2
+ C
f. For the integral below, we make the substitution u = - ( x
2
- 2 x ), so du = - (2 x - 2) dx .
It follows that:
Z
x - 1
e x 2 - 2 x
dx = -
1
2
Z
e
- ( x
2
- 2 x )
( - 2( x - 1)) dx
= -
1
2
Z
e
u
du
= -
1
2
e
u
+ C
= -
1
2
e
- x
2
+2 x
+ C
h. For the integral below, we make the substitution u = cos(2 x ), so du = - 2 sin(2 x ) dx . It
follows that:
Z
sin(2 x )
cos(2 x )
dx = -
1
2
Z
1
cos(2 x )
( - 2 sin(2 x )) dx
= -
1
2
Z
1
u
du
= -
1
2
ln | u | + C
= -
1
2
ln | cos(2 x ) | + C
,

i. For the integral below, we make the substitution u = x
1
2
, so du =
1
2
x
-
1
2
dx . It follows
that:
Z
e
√
x
√
x
, dx = 2
Z
e
√
x
1
2
x
-
1
2
dx
= 2
Z
e
u
du
= 2 e
u
+ C
= 2 e
√
x
+ C
2. Solutions for the differential equations.