So by P2 we write
2
3
1
x–
x dx
−∫
 =
0
1
2
3
3
3
1
0
1
(
– )
– (
– )
(
– )
x
x dx
x
x dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
3
3
3
1
0
1
(
– )
( –
)
(
– )
x
x dx
x
x
dx
x
x dx
−
+
+
∫
∫
∫
=
0
1
2
4
2
2
4
4
2
–1
0
1
–
–
–
4
2
2
4
4
2
x
x
x
x
x
x






+
+












=
(
)
1
1
1
1
1
1
–
–
–
4 – 2 –
–
4
2
2
4
4
2






+
+












=
1
1
1
1
1
1
–
2
4
2
2
4
4
2
+
+
−
+
−
+
 = 3
3
11
2
2
4
4
−
+
=
Example 31 Evaluate 
2
–4
4
sin x dx
π
π
∫
Solution We observe that sin2 x is an even function Therefore, by P7 (i), we get
2
–4
4
sin x dx
π
∫π
 =
2
204
sin x dx
∫π
344
MATHEMATICS
=
4
0
(1 cos 2 )
2
2
x dx
π
−
∫
 = 
4
0 (1
cos 2 )x dx
π
−
∫
=
4
0
–1
sin 2
2
x
x
π






 = 
1
1
–
sin
–0
–
4
2
2
4
2
π
π
π


=




Example 32 Evaluate 
2
0
sin
1
xcos
x
dx
x
π
+
∫
Solution Let I = 
2
0
1sin
xcos
x
dx
x
π
+
∫ 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