)
Then   sin 2 cos 3
∫
x
xdx  = 
1
sin 5
sin
2
•




∫
∫
x dx
x dx
= 
1
1 cos 5
cos
C
2
–5
x
x


+
+




= 
1
1
cos 5
cos
C
10
2
–
x
x
+
+
(iii)
From the identity sin 3x = 3 sin x – 4 sin3 x, we find that
sin3x = 3sin
sin 3
4
x –
x
Therefore,      
3
∫sin x dx
 = 3
1
sin
sin 3
4
4
x dx –
x dx
∫
∫
                                      = 
3
1
–
cos
cos 3
C
4
12
x
x
+
+
Alternatively, 
3
2
sin
sin
sin
x dx
x
x dx
=
∫
∫
 = 
(1– cos2
x) sin
x dx
∫
Put cos x = t so that – sin x dx = dt
Therefore,     
3
∫sin x dx
 = 
(
)
1 – t2
dt
−∫
 = 
3
2
C
t3
–
dt
t dt
– t
+
=
+
+
∫
∫
= 
3
1
cos
cos
C
3
–
x
x
+
+
Remark It can be shown using trigonometric identities that both answers are equivalent INTEGRALS         307
EXERCISE 7 3
Find the integrals of the functions in Exercises 1 to 22:
1 sin2 (2x + 5)
2