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

Solution:

We have, 

${\int }_0^{1/2} \frac{x^2}{{\left(1-x^2\right)}^{3/2}}dx=\frac{k}{6}$

$\Rightarrow {\int }_0^{\pi /6} \frac{{\sin }^2\theta }{{\cos }^3\theta }\cos \theta d\theta =\frac{k}{6}$, where $x=\sin \theta$

$\begin{array}{l}\Rightarrow {\int }_0^{\pi /6} {\tan }^2\theta d\theta =\frac{k}{6}\\ \Rightarrow {\int }_0^{\pi /6} \left({\sec }^2\theta -1\right)d\theta =\frac{k}{6}\\ \Rightarrow [\tan \theta -\theta {]}_0^{\pi /6}=\frac{k}{6}\Rightarrow \frac{1}{\sqrt{3}}-\frac{\pi }{6}=\frac{k}{6}\Rightarrow k=2\sqrt{3}-\pi .\end{array}$