(1)
Then, by P4
I =
4
2
0
4
4
sin (
)
2
sin (
)
cos (
)
2
2
x
dx
x
x
π
π −
π
π
−
+
−
∫
 = 
4
2
4
4
0
coscos
sin
x
dx
x
x
π
+
∫ (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
=
=
−
=
∫