
Methods of Integration 12-31
Example 12.41 Evaluate
dx
x2 3 1sin
.
−
∫
Solution: Here a = -1, b = 2 and n = 3. Also a < b.
So we apply the formula
dx
a b nx
n b a
a
nx
b
b a
+
∫
= −
−
+
−
−
sin
tanh
tan
2
2
2 2
1
2 2
+
−
∫
= −
− −
− +
− −
−
c
dx
x
x
2 3 1
2
3 2 1
3
2
2
2 1
2 2
1
2 2
sin
( ) ( )
tanh
tan
( )
+
=
−
+
−
c
x
c
2
3 3
3
2
2
3
1
tanh
tan
.
Example 12.42 Evaluate
dx
x5 3+
∫
sin
.
Solution: Here a = 5, b = 3 and n = 1. Also a > b.
So, we have
dx
a b nx
n a b
a
nx
b
a b
+
∫
=
−
+
−
+
−
sin
tan
tan
2
2
2 2
1
2 2
cc
dx
x
x
5 3
2
1 5 3
1
5 3
5
2
3
2 2
1
2 2
+
∫
=
− −
+
−
sin
( )
tan tan
+
=
+
+
−
c
x
c
1
2
5
2
3
4
1
tan
tan
.
Evaluation of
dx
a b nx c nx+ +
∫
sin ...