Question

Show that ${\int }_0^1\frac{\log x}{\sqrt{(1-x^2)}}dx=\frac{\pi }{2}.\log \frac{1}{2}.$

Answer 1

Let $I={\int }_0^1\frac{\log x}{\sqrt{(1-x^2)}}dx$
Put $x=\sin t$

$\Rightarrow dx=\cos tdt$

$I={\int }_0^{\frac{\pi }{2}}\log \sin tdt$ ....(1)
Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
$\therefore I={\int }_0^{\pi /2}\log \sin (\frac{\pi }{2}-t)dt$

$\Rightarrow I={\int }_0^{\pi /2}\log \cos tdt$ ....(2)

Adding (1) and (2), we get

$\therefore 2I={\int }_0^{\pi /2}(\log \sin t+\log \cos t)dt$

$\Rightarrow 2I={\int }_0^{\pi /2}\log \left(\sin t\cos t\right)dt$

$={\int }_0^{\pi /2}\log \frac{\sin 2t}{2}dt$

$\Rightarrow I={\int }_0^{\pi /2}\log \sin 2tdt-{\int }_0^{\pi /2}\log 2dt$

$\Rightarrow I={\int }_0^{\pi /2}\log \sin 2tdt-\frac{\pi }{2}\log 2$
Put $2t=u\Rightarrow 2dt=du$

$\therefore 2I=\frac{1}{2}{\int }_0^{\pi }\log \sin udu-\frac{\pi }{2}\log 2$ $(\because f(2a-x)=f\left(x\right)$

$2I=\frac{1}{2}.2{\int }_0^{\pi /2}\log \sin udu-\frac{\pi }{2}\log 2$

$\Rightarrow 2I={\int }_0^{\pi /2}\log \sin udu-\frac{\pi }{2}\log 2$
$2I=I+\frac{\pi }{2}\log \frac{1}{2}$
$\therefore I={\int }_0^{\pi /2}\log \sin tdt={\int }_0^{\pi /2}\log \cos tdt=\frac{\pi }{2}\log \frac{1}{2}$