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
= Thus, we use the substitution
1
so that
giving
2
x
t
dx
dt
x
=
=
 dx = 2t dt Thus,
4
2
4
2
tan
sec
2 tan
sec
x
x
t
t
t dt
dx
t
x
=
∫
∫
 = 
4
2
2 tan
tsec
t dt
∫
Again, we make another substitution tan t = u so that
sec2 t dt = du
Therefore,
4
2
4
2 tan
sec
2
t
t dt
u du
=
∫
∫
 = 
5
2
C
5
u +
=
2 tan5
C
5
t +
 (since u = tan t)
=
2 tan5
C (since
)
5
x
t
x
+
=
Hence,
4
2
tan
xsec
x dx
x
∫
 =
2 tan5
C
5
x +
Alternatively, make the substitution tan
x
t
=
(iv)
Derivative of  
1
2
1
tan
1
– x
x
= +