Question

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

Answer 1

Using the property,

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

Hence we can write $I={\int }_0^{\frac{\pi }{2}}\log \sin xdx={\int }_0^{\frac{\pi }{2}}\log \sin (\frac{\pi }{2}-x)dx$

$={\int }_0^{\frac{\pi }{2}}\log \cos xdx$

$I={\int }_0^{\frac{\pi }{2}}\log \sin xdx={\int }_0^{\frac{\pi }{2}}\log \cos xdx$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \sin xdx+{\int }_0^{\frac{\pi }{2}}\log \cos xdx$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log \sin x+\log \cos xdx$

$2I={\int }_0\frac{\pi }{2}\left(\log \left(\sin x\cos x\right)\right)dx$

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

$={\int }_0^{\frac{\pi }{2}}(\log \sin 2x-\log 2)dx$

${\int }_0^{\frac{\pi }{2}}\log \sin 2xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx$

$={\int }_0^{\frac{\pi }{2}}\log \sin 2xdx-\frac{\pi }{2}\log 2$ ........$\left(1\right)$

Let $I_1={\int }_0^{\frac{\pi }{2}}\log \sin 2xdx$ and $t=2x$ then $I_1=\frac{1}{2}{\int }_0^{\pi }\log \sin tdt$

and using the property ${\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f(2a-x)dx$ if $f(2a-x)=f\left(x\right)$

Note that $\log \sin t=\log \sin (\pi -t)$ and we get

$I_1=\frac{1}{2}{\int }_0^{\pi }\log \sin tdt={\int }_0^{\frac{\pi }{2}}\log \sin tdt=I$

Hence $\left(1\right)$ becomes $2I=I-\frac{\pi \log 2}{2}$

or $I=\frac{-\pi \log 2}{2}$