The value of 
02
4
3sin
log
4
3cos
x
dx
x
π


+


+


∫
 is
(A) 2
(B)
43
(C) 0
(D) –2
348
MATHEMATICS
Miscellaneous Examples
Example 37 Find cos 6
1 sin 6
x
x dx
+
∫
Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx
Therefore
21
1
cos 6
1 sin 6
6
x
x dx
t dt
+
=
∫
∫
=
3
3
2
2
1
2
1
( )
C = 
(1 sin 6 )
C
6
3
9
t
x
×
+
+
+
Example 38 Find 
1
4
4
5
(
)
x
x
dx
−x
∫
Solution We have 
1
1
4
4
4
3
5
4
1
(1
)
(
)
x
x
x
dx
dx
x
x
−
−
=
∫
∫
Put 
– 3
3
4
1
3
1
1–
, so that
x
t
dx
dt
x
x
−
=
=
=
Therefore 
1
1
4
4
4
5
(
)
31
x
x
dx
t
dt
−x
=
∫
∫
 = 
5
5
4
4
3
1
4
4
1
C =
1
C
3
5
15
t
x


×
+
−
+




Example 39 Find 
4
2
(
1) (
1)
x dx
x
x
−
+
∫
Solution We have
4
2
(
1)(
1)
x
x
x
−
+
 =
3
12
(
1)
1
x
x
x
x
+
+
−
+
−
=
12
(
1)
(
1) (
1)
x
x
x
+
+
−
+ (1)
Now express
2
1
(
1)(
1)
x
x
−
+
 =
2
A
B
C
(
1)
(
1)
x
x
x
+
+
−
+ (2)
INTEGRALS         349
So
1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give 
1
1
A
, B
C
–
2
2
=
=
= Substituting values of A, B and C in (2), we get
2
1
(
1) (
1)
x
x
−
+
 =
2
2
1
1
1
2(
1)
2 (
1)
2(
1)
x
x
x
x
−
−
−
+
+