Then, by P4, we have
I =  
2
0
(
)sin (
)
1
cos (
)
x
x dx
x
π π −
π −
+
π −
∫
=
2
0
(
)sin
1 cos
x
x dx
x
π π −
+
∫
 = 
2
0
sin
I
1
cos
x dx
x
π
π
−
+
∫
or
2 I =
2
0
1sin
cos
x dx
x
π
π
+
∫
or
I =
2
0
sin
2
1
cos
x dx
x
π
π
+
∫
Put cos x = t so that – sin x dx = dt 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