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