Question

Integrate:
${\int }_0^{\pi /2}\log \left(\sin x\right)dx$

Answer 1

$I={\int }_0^{\pi /2}log\left(sinx\right)dx={\int }_0^{\pi /2}logsin(\frac{\pi }{2}-x)dx={\int }_0^{\pi /2}logcosxdx$

$[\therefore {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx]$

$\Rightarrow 2I={\int }_0^{\pi /2}(logsinx+logcosx)dx$

$={\int }_0^{\pi /2}(logsinx+logcosx+log2-log2)dx$

$={\int }_0^{\pi /2}log\left(sin2x\right)dx-{\int }_0^{\pi /2}log2dx$

Let $2x=t\Rightarrow 2dx=dt.$ When $x=0,t=0$ & $x=\pi /2,t=\pi$

$\Rightarrow 2I=\frac{1}{2}{\int }_0^{\pi }log\left(sint\right)dt-\frac{\pi }{2}log2$

$=\frac{2}{2}{\int }_0^{\pi /2}log\left(sint\right)dt-\frac{\pi }{2}log2$ $[\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$ when $f(2a-x)=f\left(x\right)]$

$={\int }_0^{\pi /2}logsinxdx-\frac{\pi }{2}log2$ (changing t by x)

$=I-\frac{\pi }{2}log2$

$\Rightarrow I=-\frac{\pi }{2}log2$