Question

Prove that ${\int }_0^{\pi /2}$ $ln\left(\sin x\right)dx={\int }_0^{\pi /2}ln\left(cosx\right)dx={\int }_0^{\pi /2}ln\left(sin2x\right)dx=-\frac{\pi }{2}.ln2$.

Answer 1

Let $S={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)dx$

Put $t=\frac{\pi }{2}-x$

$S=-{\int }_0^{\frac{\pi }{2}}\ln \left(\sin (\frac{\pi }{2}-t)\right)dt={\int }_0^{\frac{\pi }{2}}\ln \left(\cos t\right)dt$

Sum the two expressions, we get

$2S={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\right)dx+{\int }_0^{\frac{\pi }{2}}\ln \left(\cos x\right)dx$

$2S={\int }_0^{\frac{\pi }{2}}\ln \left(\sin x\cos x\right)dx={\int }_0^{\frac{\pi }{2}}ln\left(\frac{1}{2}\sin \left(2x\right)\right)dx=\frac{\pi }{2}\ln \left(\frac{1}{2}\right)+{\int }_0^{\frac{\pi }{2}}\ln \sin \left(2x\right)dx$

Let us integrate the last integral by substituting $t=2x$ and $t=2x-\frac{\pi }{2}$ in two intervals.

So, ${\int }_0^{\frac{\pi }{2}}\ln \left(\sin \left(2x\right)\right)dx={\int }_0^{\frac{\pi }{4}}\ln \left(\sin \left(2x\right)\right)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\ln \left(\sin \left(2x\right)\right)dx$

So,

$2S=\frac{\pi }{2}\ln \left(\frac{1}{2}\right)+S⟹S=\frac{\pi }{2}\ln \left(\frac{1}{2}\right)=-\frac{\pi }{2}\ln \left(2\right)$

Also, $S={\int }_0^{\frac{\pi }{2}}\ln \left(\sin \left(2x\right)\right)dx$