Mathematics - Definite Integration Question with Solution | TestHub

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

Let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$

$\Rightarrow \frac{k}{6}={\int }_0^{\frac{\pi }{6}}\frac{{\sin }^2 \theta }{{\left(1-{\sin }^2 \theta \right)}^{{\displaystyle \frac{3}{2}}}}\cos \theta d\theta \Rightarrow \frac{k}{6}={\int }_0^{\frac{\pi }{6}}\frac{{\sin }^2 \theta }{\left({\cos }^3 \theta \right)}\cos \theta d\theta$

$\Rightarrow \frac{k}{6}={\int }_0^{\frac{\pi }{6}}{\tan }^2\theta d\theta ={\int }_0^{\frac{\pi }{6}}\left(se{c}^2\theta -1\right)d\theta \Rightarrow \frac{k}{6}={\left(\tan \theta -\theta \right)}_0^{\frac{\mathrm{\pi }}{6}}=\left(\frac{1}{\sqrt{3}}-\frac{\pi }{6}\right)=\frac{2\sqrt{3}-\pi }{6}$

$\Rightarrow k=2\sqrt{3}-\pi$