Example 15 Find (
)
3 sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
Solution Let y = sinφ
Then
dy = cosφ dφ
INTEGRALS         321
Therefore,
(
)
3sin2
2 cos
5
cos
4sin
–
d
–
–
φ
φ
φ
φ
φ
∫
 =
2
(3 – 2)
5
(1
)
4
y
dy
–
– y
–
y
∫
=
32
2
4
4
y –
dy
y – y +
∫
=
(
)
2
3
2
I (say)
2
y –
y –
=
∫
Now, we write
(
)
2
3
2
2
y –
y –
 =
2
A
B
2
(
2)
y
+y
−
−
[by Table 7 2 (2)]
Therefore,
3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4 Therefore, the required integral is given by
I =
2
3
4
[
+
]
2
(
2)
dy
y –
y –
∫
 = 
2
3
2+ 4
(
2)
dy
dy
y –
y –
∫
∫
=
1
3log
2
4
C
2
y
–
y


−
+
+


−


=
4
3log sin
2
C
2
sin
–
φ −
+
+
φ
=
4
3log (2
sin )
+ C
2
sin
−
φ +
−
φ
 (since, 2 – sinφ is always positive)
Example 16 Find 
2
2
1
(
2) (
1)
x
x
dx
x
x
+
+
+
+
∫
Solution The integrand is a proper rational function Decompose the rational function
into partial fraction [Table 2