(2)
Adding (1) and (2), we get
2I =
4
4
2
2
2
4
4
0
0
0
sin
cos
[ ]
2
sin
xcos
x dx
dx
x
x
x
π
π
π
+
π
=
=
=
+
∫
∫
Hence
I =
4
π
Example 35 Evaluate 
3
6 1
tan
dx
x
π
π
+
∫
Solution  Let I = 
3
3
6
6
cos
1
tan
cos
sin
x dx
dx
x
x
x
π
π
π
π
=
+
+
∫
∫ (1)
Then, by P3
I =
3
6
cos
3
6
cos
sin
3
6
3
6
x
dx
x
x
π
π
π
π


+
−




π
π
π
π




+
−
+
+
−








∫
=
3
6
sin
sin
cos
x
dx
x
x
π
π
+
∫ (2)
Adding (1) and (2), we get
2I =
[ ]
3
3
6
6
3
6
6
dx
x
π
π
π
π
π
π
π
=
=
−
=
∫ Hence I
12
=π
346
MATHEMATICS
Example 36 Evaluate 
2
0 log sin x dx
π
∫
Solution Let I =
2
0 logsinx dx
π
∫
Then, by P4
I =
2
2
0
0
log sin
log cos
2
x dx
x dx
π
π
π

−
=




∫
∫
Adding the two values of I, we get
2I =
(
)
02
log sin
logcos
x
x dx
π
+
∫
=
(
)
02
log sin
cos
log2
log2
x
x
dx
π
+
−
∫
(by adding and subtracting log2)
=
2
2
0
0
log sin2
log2
x dx
dx
π
π
−
∫
∫
(Why