Question

Value of ${\int }_0^1\frac{dx}{(1+x^2)\sqrt{1-x^2}}$ is?

• A. $\frac{\pi }{2\sqrt{2}}$
• B. $\frac{\pi }{\sqrt{2}}$
• C. $\sqrt{2}\pi$
• D. $2\sqrt{2\pi }$

Answer 1

Answer: (A)

$\frac{\pi }{2\sqrt{2}}$

Let $I=\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$
Substitute $x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$
$I=\int \frac{-tdt}{(t^2+1)\sqrt{t^2-1}}$
Substitute $t^2-1=u^2\Rightarrow 2tdt=2udu$
$I=-\int \frac{udu}{(u^2+1)u}=-\int \frac{1}{u^2+{\left(\sqrt{2}\right)}^2}du$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)=-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{2}x}{\sqrt{1-x^2}}\right)$
Therefore ${\int }_0^{1}Idx=\frac{\pi }{2\sqrt{2}}$