When x = 0, t = 1 and when x = π, t = – 1 Therefore, (by P1) we get
I =
1
2
1
–
2
1
dt
t
−
π
+
∫
= 
1
2
1
2
1
dt
t
−
π
+
∫
=
1
2
0 1
dt
t
π
+
∫
 (by P7, 
2
1
since
1 t
+
 is even function)
=
2
1
– 1
– 1
1
0
tan
tan
1– tan
0
– 0
4
4
t
−
π
π






π
= π
= π
=








Example 33 Evaluate 
1
5
4
1 sin
xcos
x dx
−∫
Solution Let I = 
1
5
4
1sin
xcos
x dx
−∫ Let f(x) = sin5 x cos4 x Then
f (– x) = sin5 (– x) cos4 (– x) = – sin5 x cos4 x = – f (x), i