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 These will be used later without
reference (i) ∫tan
= log sec
+ C
x dx
x
We have
sin
tan
cos
x
x dx
dx
x
=
∫
∫
302
MATHEMATICS
Put  cos x = t so that sin x dx = – dt
Then
tan
log
C
log cos
C
dt
x dx
–
–
t
–
x
t
=
=
+
=
+
∫
∫
or
tan
log sec
C
x dx
x
=
+
∫
(ii) ∫cot
= log sin
+ C
x dx
x
We have
cos
cot
sin
x
x dx
dx
x
=
∫
∫
Put  sin x = t so that cos x dx = dt
Then
cot
dt
x dx
t
=
∫
∫
 = log
t +C
 = log sin
C
x +
(iii) ∫sec
= log sec
+ tan
+ C
x dx
x
x
We have
sec
(sec
tan )
sec
sec
+ tan
x
x
x
x dx
dx
x
+x
=
∫
∫
Put sec x + tan x = t so that sec x (tan x + sec x) dx = dt
Therefore, sec
log
+ C = log sec
tan
C
dt
x dx
t
x
x
t
=
=
+
+
∫
∫
(iv) ∫cosec
= log cosec
– cot
+ C
x dx
x
x
We have
cosec
(cosec
cot )
cosec
(cosec
cot )
x
x
x
x dx
dx
x
+x
=
+
∫
∫
Put cosec x + cot x = t so that – cosec x (cosec x + cot x) dx = dt
So
cosec
–
–log| |
– log|cosec
cot
|
C
dt
x dx
t
x
x
t
=
=
=
+
+
∫
∫
=
2
2
cosec
cot
– log
C
cosec
cot
x
x
x
x
−
+
−
= log cosec
cot
C
x –
x +
Example 6 Find the following integrals:
(i)
3
2
sin
xcos
x dx
∫
(ii)    
sin
sin (
)
x
dx
x
+a
∫
    (iii)  
1
1
tan
dx
x
∫+
INTEGRALS         303
Solution
(i)
We have
3
2
2
2
sin
cos
sin
cos
(sin )
x
x dx
x
x
x dx
=
∫
∫
= 
2
2
(1– cos
) cos
(sin )
x
x
x dx
∫
Put t = cos x so that dt = – sin x dx
Therefore,    
2
2
sin
cos
(sin )
x
x
x dx
∫
 = 
2
2
(1–
t)
t dt
−∫
= 
3
5
2
4
(
–
)
C
3
5
t
t
–
t
t
dt
–
–


=
+




∫
= 
3
5
1
1
cos
cos
C
3
5
–
x
x
+
+
(ii)
Put x + a = t Then dx = dt