Question

If the value of the integral ${\int }_0^{\frac{1}{2}}\frac{x^2}{{(1-x^2)}^{\frac{3}{2}}}dx$ is $\frac{k}{6}$, then $k$ is equal to:

• A. $2\sqrt{3}+\pi$
• B. $3\sqrt{2}+\pi$
• C. $3\sqrt{2}-\pi$
• D. $2\sqrt{3}-\pi$

Answer 1

The correct option is D $2\sqrt{3}-\pi$

$I={\int }_0^{\frac{1}{2}}\frac{x^2}{{(1-x^2)}^{\frac{3}{2}}}dx$
Let $x=\sin \theta ;dx=\cos \theta d\theta$
$={\int }_0^{\frac{\pi }{6}}\frac{{\sin }^2\theta }{{\cos }^3\theta }\cdot \cos \theta d\theta$
$={\int }_0^{\frac{\pi }{6}}{\tan }^2\theta d\theta ={[\tan \theta -\theta ]}_0^{\frac{\pi }{6}}$
$\Rightarrow (\frac{1}{\sqrt{3}}-\frac{\pi }{6})=\frac{k}{6}$
$\Rightarrow \frac{2\sqrt{3}-\pi }{6}=\frac{k}{6}$
$\therefore k=2\sqrt{3}-\pi$