Put 
23
3
30 – Then –
2
x
t
x dx
dt
=
=
 or 
2
–
3
x dx
dt
=
Thus,  
3
2
2
2
–2
3
(30 –
)
x
dt
dx
t
x
=
∫
∫
 = 
2 1
3 t
 
 
  = 
3
2
2
1
F ( )
3 (30 –
)
x
x



 =






INTEGRALS         337
Therefore, by the second fundamental theorem of calculus, we have
I =
9
3
2
4
2
1
F(9) – F(4)
3
(30 –
)
x




=






=
2
1
1
3 (30 – 27)
30 –8


−




 = 
2 1
1
19
3 3
22
99


−
=




(iii)
Let 
2
1
I
(
1) (
2)
x dx
x
x
=
+
+
∫
Using partial fraction, we get  
–1
2
(
1) (
2)
1
2
x
x
x
x
x
=
+
+
+
+
+
So
(
1) (
2)
x dx
x
x
+
+
∫
 = – log
1
2log
2
F( )
x
x
x
+
+
+
=
Therefore, by the second fundamental theorem of calculus, we have
I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]
= – 3 log 3 + log 2 + 2 log 4 = 
32
log
27






(iv)
Let 
3
04
I
sin 2 cos2
t
t dt
= ∫π Consider 
3
sin 2 cos2
t
t dt
∫
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt = 1
2  du
So
3
sin 2 cos2
t
t dt
∫
 =
3
1
2
u du
∫
=
4
4
1
1
[
]
sin 2
F ( )say
8
8
u
t
t
=
=
Therefore, by the second fundamental theorem of integral calculus
I =
4
4
1
1
F ( ) – F (0)
[sin
–sin 0]
4
8
2
8
π
π
=
=
338
MATHEMATICS
EXERCISE 7 9
Evaluate the definite integrals in Exercises 1 to 20