Question

Integral of log (sin^2x)

Answer 1

Integral of Log Sin² x =  2 * I  where

I = Integral  Log Sin x            with limits from  x = 0 to π/2

$I= \int \limits_{0}^{\pi/2} {Log\ Sinx} \, dx\\\\y=\pi/2-x,\ \ dy=-dx\\\\\implies I= \int \limits_{0}^{\pi/2} {Log\ Cosy} \, dy\\\\I+I= \int \limits_{0}^{\pi /2} {Log(sinx\ cosx)} \, dx\\\\2I=\int \limits_{0}^{\pi/2} {(Log\ Sin2x-Log2)} \, dx\\\\2I=\frac{1}{2}\int \limits^{\pi}_{0} {Log\ sin\ z} \, dz- \frac{\pi}{2} Log2\\\\2I=\frac{1}{2}*2I-\frac{\pi}{2}Log2\\\\I=\frac{\pi}{2}Log2$

Answer = - π Log 2.