Then
I =
(
)
– 1
– 1
2
2
1
2
2 tan
C =
2 tan
C
2
2
2
t
dy
y
t
y


−



=
+
+
+
∫
=
2
– 1
– 1
1
tan
1
2 tan
C =
2 tan
C
2
2tan
t
x
t
x

−
−


+
+










Example 42 Find 
4
sin 2 cos 2
9– cos (2 )
x
x dx
x
∫
Solution Let 
4
sin 2 cos 2
I
9 – cos 2
x
x dx
x
=∫
INTEGRALS         351
Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt
Therefore
–1
1
2
2
1
1
1
1
I
–
–
sin
C
sin
cos 2
C
4
4
3
4
3
9 –
dt
t
x
t
−




=
=
+
= −
+








∫
Example 43 Evaluate 
123
sin (
)
x
x dx
−
π
∫
Solution Here f (x) = | x sin πx | = 
sin
for
1
1
3
sin
for 1
2
x
x
x
x
x
x
π
− ≤
≤
−
π
≤
≤

Therefore
23
1 |
sin
|
x
x dx
−
π
∫
 =
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
+
−
π
∫
∫
=
3
1
2
1
1
sin
sin
x
x dx
x
x dx
−
π
−
π
∫
∫
Integrating both integrals on righthand side, we get
23
1 |
sin
|
x
x dx
−
π
∫
 =
3
1
2
2
2
1
1
–
cos
sin
cos
sin
x
x
x
x
x
x
−
π
π
−
π
π




+
−
+




π
π
π
π




=
2
2
2
1
1
3
1


− −
−
=
+


π
π
π
π
π


Example 44 Evaluate 
2
2
2
2
0
cos
sin
x dx
a
x
b
x
π
+
∫
Solution Let I = 
2
2
2
2
2
2
2
2
0
0
(
)
cos
sin
cos (
)
sin (
)
x dx
x dx
a
x
b
x
a
x
b
x
π
π
π −
=
+
π −
+
π −
∫
∫
(using P4)
=
2
2
2
2
2
2
2
2
0
0
cos
sin
cos
sin
dx
x dx
a
x
b
x
a
x
b
x
π
π
π
−
+
+
∫
∫
=
2
2
2
2
0
I
cos
sin
dx
a
x
b
x
π
π
−
+
∫
Thus
2I =
2
2
2
2
0
cos
sin
dx
a
x
b
x
π
π
+
∫
352
MATHEMATICS
or
I =
2
2
2
2
2
2
2
2
2
0
20
2
2
cos
sin
cos
sin
dx
dx
a
x
b
x
a
x
b
x
π
π
π
=π
⋅
+
+
∫
∫
(using P6)
=
2
4
2
2
2
2
2
2
2
2
0
4
cos
sin
cos
sin
π
π
π


π
+


+
+




∫
∫
dx
dx
a
x
b
x
a
x
b
x
= 
2
2
2
4
2
2
2
2
2
2
0
4
sec
cosec
tan
cot
π
π
π


π
+


+
+




∫
∫
xdx
xdx
a
b
x
a
x
b
= 
(
)
0
1
2
2 2
2
2
2
0
1
tan
t
cot


π
−
=
=


+
+


∫
∫
dt
du
put
x
and
x
u
a
b t
a u
b
= 
1
0
–1
–1
0
1
tan
–
tan
π
π












bt
au
ab
a
ab
b
 =  
–1
–1
tan
tan
π 

+




b
a
ab
a
b = 
2
2
π
ab
Miscellaneous Exercise on Chapter 7
Integrate the functions in Exercises 1 to 24 1 3
1
x
−x
2 1
x
a
x
b
+
+
+
3