Question

If ${\int }_0^1\frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}}=\frac{\pi A}{\sqrt{6}(\sqrt{3}+1)\left(157\right)}$, then $A$ is equal to

Answer 1

Since $x(1-x)=\frac{1}{4}-{(x-\frac{1}{2})}^2$,

so put $x=\frac{1}{2}+\frac{1}{2}\sin \theta$

Therefore,
${\int }_0^1\frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}}$

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\frac{1}{2}\cos \theta }{\left(\frac{1}{4}\right(1+\sin \theta {)}^2+\frac{3}{2}(1+\sin \theta )+2)\frac{1}{2}\cos \theta }d\theta$

$=4{\int }_0^{\pi }\frac{d\varphi }{{\cos }^2\varphi -8\cos \varphi +15}$, where $\varphi =\theta +\frac{\pi }{2}$

$=\frac{4}{2}\left[{\int }_0^{\pi }\frac{d\varphi }{\cos \varphi -(4+1)}{\int }_0^{\pi }\frac{d\varphi }{\cos \varphi -(4-1)}\right]$

Let, $I_1={\int }_0^{\pi }\frac{d\varphi }{\cos \varphi -a}$, where $a>1$

Put $\tan \frac{\varphi }{2}=t\Rightarrow d\varphi =2\frac{dt}{(1+t^2)}$

$\therefore I_1={\int }_0^{\infty }\frac{\frac{2}{(1+t^2)}}{\frac{(1-t^2)}{(1+t^2)}-a}dt=\frac{-2}{1+a}{\int }_0^{\infty }\frac{dt}{t^2+\frac{(a-1)}{(a+1)}}$

$=\frac{-2}{1+a}\sqrt{\frac{a+1}{a-1}}{\left({\tan }^{-1}\frac{t\sqrt{a+1}}{\sqrt{a-1}})\right]}_0^{\infty }$

$=\frac{-2}{\sqrt{a^2-1}}\cdot \frac{\pi }{2}=\frac{-\pi }{\sqrt{a^2-1}}$

Thus,

$I=2[-\frac{\pi }{\sqrt{24}}+\frac{\pi }{\sqrt{8}}]$

$=\pi (\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}})=\pi \frac{(\sqrt{3}-1)}{\sqrt{6}}$

Hence $A=314$