(1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
0
0
0
( )
( )
( )
2
( )
a
a
a
a
a f x dx
f x dx
f x dx
f x dx
−
=
+
=
∫
∫
∫
∫
(ii) If f is an odd function, then f (–x) = – f(x) and so (1) becomes
0
0
( )
( )
( )
0
a
a
a
a f x dx
f x dx
f x dx
−
= −
+
=
∫
∫
∫
Example 30 Evaluate 
2
3
1
x–
x dx
−∫
Solution We note that x3 – x ≥ 0 on [– 1, 0] and x3 – x ≤ 0 on [0, 1] and that
x3 – x ≥ 0 on [1, 2] 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