Hence I
12
=π
346
MATHEMATICS
Example 36 Evaluate 
2
0 log sin x dx
π
∫
Solution Let I =
2
0 logsinx dx
π
∫
Then, by P4
I =
2
2
0
0
log sin
log cos
2
x dx
x dx
π
π
π

−
=




∫
∫
Adding the two values of I, we get
2I =
(
)
02
log sin
logcos
x
x dx
π
+
∫
=
(
)
02
log sin
cos
log2
log2
x
x
dx
π
+
−
∫
(by adding and subtracting log2)
=
2
2
0
0
log sin2
log2
x dx
dx
π
π
−
∫
∫
(Why )
Put 2x = t in the first integral Then 2 dx = dt, when x = 0, t = 0 and when 
2
x
=π
,
t = π Therefore
2I =
0
1
log sin
log2
2
2
t dt
π
−π
∫
=
02
2
log sin
log2
2
2
t dt
π
−π
∫
 [by P6 as sin (π – t) = sin t)
=
2
0 log sin
log2
2
x dx
π
−π
∫
 (by changing variable t to x)
= I
log2
π2
−
Hence
2
0 log sin x dx
∫π
 = –
2log2
π