Question

If $I={\int }_0^1\frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}}$ then $I$ equals

• A. $2\pi$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{\sqrt{6}}(\sqrt{3}-1)$

Answer 1

The correct option is D $\frac{\pi }{\sqrt{6}}(\sqrt{3}-1)$

Substitute $x={\sin }^2\theta$, so that

$I={\int }_0^{\pi /2}\frac{2\sin \theta \cos \theta }{(1+{\sin }^2\theta )(2+{\sin }^2\theta )\sin \theta \cos \theta }d\theta$

$=2{\int }_0^{\pi /2}\frac{{\sec }^4\theta }{({\sec }^2\theta +{\tan }^2\theta )(2+{\sec }^2\theta +{\tan }^2\theta )}d\theta$

Substitute $\tan \theta =t,$ so that

$I=2{\int }_0^{\infty }\frac{(1+t^2)dt}{(1+2t^2)(2+3t^2)}$

$=2{\int }_0^{\infty }[\frac{1}{1+2t^2}-\frac{1}{2+3t^2}]dt$

$=2{\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\right(\sqrt{2}t)-\frac{1}{\sqrt{2}\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}t}{\sqrt{2}}\right)]}_0^{\infty }$

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