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
= + Thus, we use the substitution
tan–1 x = t so that 
2
1
dx
x
+
 = dt Therefore ,  
1
sin (tan2
)
sin
1
– x dx
t dt
x
=
+
∫
∫
 =  – cos t + C = – cos (tan –1x) + C
Now, we discuss some important integrals involving trigonometric functions and
their standard integrals using substitution technique