let I = ₀ò^p q ln sin q dq

on aplying property of definite integral

I =  ₀ò^p (π-q) ln sin q dq

so 2 I =  ₀ò^p π ln sin q dq 

or I = π/2  ₀ò^p  ln sin q dq

or I = 2π/2  ₀ò^p/2 ln sin q dq    this is due to property

similarly solve  ₀ò^p q² ln sin q dq  and   then ₀ò^p q³ ln sin q dq 

finally solve the right hand side of the equation to prove LHS = RHS