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 INTEGRALS         347
EXERCISE 7 11
By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19