Thus, we make the substitution
mx = t so that mdx = dt Therefore,      
1
sin
sin
mx dx
t dt
m
=
∫
∫
 =  – 1
m
cos t + C  = – 1
m cos mx + C
INTEGRALS         301
(ii)
Derivative of x2 + 1 is 2x Thus, we use the substitution x2 + 1 = t so that
2x dx = dt Therefore,  
2 sin (2
1)
sin
x
x
dx
t dt
+
=
∫
∫
 =  – cos t + C  = – cos (x2 + 1) + C
(iii)
Derivative of 
x  is 
21
1
1
2
2
–
x
x
=