x:
(i)
sin mx
(ii)
2x sin (x2 + 1)
(iii)
4
2
tan
xsec
x
x
(iv)
1
2
sin (tan
)
1
x– x
+
Solution
(i)
We know that derivative of mx is m 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