Question

The value of ${\int }_0^{1/2}\frac{dx}{\sqrt{(x-x^2)}}$is $\frac{\pi }{k}.$ Find the value of $k$.

Answer 1

$I={\int }_0^{\frac{\pi }{2}}\frac{1}{\sqrt{x-x^2}}dx={\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{\frac{1}{4}-{(x-\frac{1}{2})}^2}}dx$
Put $t=x-\frac{1}{2}\Rightarrow dt=dx$
$I={\int }_{-\frac{1}{2}}^0\frac{1}{\sqrt{\frac{1}{4}-t^2}}dt=2{\int }_{-\frac{1}{2}}^0\frac{1}{\sqrt{1-{4t}^2}}dt$
$={\left[si{n}^{-1}2t\right]}_{-\frac{1}{2}}^0=0-si{n}^{-1}\left(\frac{-2}{2}\right)=\frac{\pi }{2}$