Then
2
2
2
(
1) (
4)
x
x
x
+
+
 = (
1) (
4)
y
y
y
+
+
Write
(
1) (
4)
y
y
y
+
+
 =
A
B
1
4
y
y
+
+
+
So that
y =  A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
A =
1
4
and B
3
3
−
=
Thus,
2
2
2
(
1) (
4)
x
x
x
+
+
 =
2
2
1
4
3(
1)
3 (
4)
–
x
x
+
+
+
Therefore,
2
2
2
(
1) (
4)
x dx
x
x
+
+
∫
 =
2
2
1
4
3
3
1
4
dx
dx
–
x
x
+
+
+
∫
∫
=
1
1
1
4
1
tan
tan
C
3
3
2
2
–
– x
–
x +
×
+
=
1
1
1
2
tan
tan
C
3
3
2
–
– x
–
x +
+
In the above example, the substitution was made only for the partial fraction part
and not for the integration part Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method 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