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 1 1
1(
1)
x
dx
−
+
∫
2