Prove that $\int_{0}^{\pi/2}$ log(sinx)dx = $\int_{0}^{\pi/2}$ log(cosx)dx = -$\pi\over 2$log 2.

Solution :

Let I = $\int_{0}^{\pi/2}$ log(sinx)dx …….(i)

then I = $\int_{0}^{\pi/2}$ $log(sin({\pi\over 2}-x))$ dx = $\int_{0}^{\pi/2}$ log(cosx)dx …….(ii)

Adding (i) and (ii), we get

2I = $\int_{0}^{\pi/2}$ log(sinx)dx + $\int_{0}^{\pi/2}$ log(cosx)dx = $\int_{0}^{\pi/2}$ (log(sinx)dx + log(cosx))

$\implies$ $\int_{0}^{\pi/2}$ log(sinxcosx)dx = $\int_{0}^{\pi/2}$ $log({2sinxcosx\over 2})$ dx

= $\int_{0}^{\pi/2}$ $log({sin2x\over 2})$ dx = $\int_{0}^{\pi/2}$ log(sin2x)dx – $\int_{0}^{\pi/2}$ log(2)dx

= $\int_{0}^{\pi/2}$ log(sin2x)dx – (log 2) ${(x)}^{\pi/2}_{0}$

$\implies$ 2I = $\int_{0}^{\pi/2}$ log(sin2x)dx – $\pi\over 2$ log 2 …(iii)

Let $I_1$ = $\int_{0}^{\pi/2}$ log(sin2x)dx, putting 2x = t, we get

$I_1$ = $\int_{0}^{\pi}$ log(sint) $dt\over 2$ = $1\over 2$ $\int_{0}^{\pi}$ log(sint)dt = $1\over 2$ 2 $\int_{0}^{\pi/2}$ log(sint)dt

$I_1$ = $\int_{0}^{\pi/2}$ log(sinx)dx

$\therefore$ (iii) becomes; 2I = I – $\pi\over 2$ log 2

Hence $\int_{0}^{\pi/2}$ log(sinx)dx = – $\pi\over 2$ log 2