(3)
Again, substituting (3) in (1), we have
4
2
(
1) (
1)
x
x
x
x
−
+
+
 =
2
2
1
1
1
(
1)
2(
1)
2 (
1)
2(
1)
x
x
x
x
x
+
+
−
−
−
+
+
Therefore
4
2
2
– 1
2
1
1
1
log
1 –
log (
1) –
tan
C
2
2
4
2
(
1) (
1)
x
x
dx
x
x
x
x
x
x
x
=
+
+
−
+
+
−
+
+
∫
Example 40 Find 
2
1
log (log )
(log )
x
dx
x


+




∫
Solution Let 
2
1
I
log (log )
(log )
x
dx
x


=
+




∫
=
2
1
log (log )
(log )
x dx
dx
x
+
∫
∫
In the first integral, let us take 1 as the second function Then integrating it by
parts, we get
I =
2
1
log (log )
log
(log )
dx
x
x
x dx
x
x
x
−
+
∫
∫
=
2
log (log )
log
(log )
dx
dx
x
x
x
x
−
+
∫
∫ (1)
Again, consider 
log
dx
x
∫
, take 1 as the second function and integrate it by parts,
we have 
2
1
1
–
–
log
log
(log )
dx
x
x
dx
x
x
x
x






= 













∫
∫ (2)
350
MATHEMATICS
Putting (2) in (1), we get
2
2
I
log (log )
log
(log )
(log )
x
dx
dx
x
x
x
x
x
=
−
−
+
∫
∫
 = 
log (log )
C
log
x
x
x
x
−
+
Example 41 Find 
cot
tan
x
x dx


+


∫
Solution We have
I =
cot
tan
x
x
dx


+


∫
tan (1
cot )
x
x dx
=
+
∫
Put tan x = t2, so that sec2 x dx = 2t dt
or
dx =  
4
2
1
t dt
t
+
Then
I =
2
4
1
2
1
(1
)
t
t
dt
t
t


+

+


∫
=
2
2
2
4
2
2
2
1
1
1
1
(
1)
2
= 2
= 2
1
1
1
2
dt
dt
t
t
t
dt
t
t
t
t
t




+
+




+






+


+
−
+








∫
∫
∫
Put 
1
t
t
−  = y, so that 
12
1
t


+


  dt = dy