Official solution
\[I = \int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{(1 - {x^2})}^{{3 \over 2}}}}}} dx\]
Let \[x = \sin \theta \]
\[ \Rightarrow dx = \cos \theta d\theta \]
When x = 0 then sin\[\theta \] = 0 \[ \Rightarrow \] \[\theta \] = 0
When \[x = {1 \over 2}\] then \[\sin \theta = {1 \over 2}\] \[ \Rightarrow \] \[\theta = {\pi \over 6}\]
\[ \therefore \] \[I = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{({{\cos }^2}\theta )}^{{3 \over 2}}}}}\cos \theta d\theta } \]
\[ = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}d\theta } \]
\[ = \int\limits_0^{{\pi \over 6}} {{{\tan }^2}d\theta } = \int\limits_0^{{\pi \over 6}} {({{\sec }^2}\theta - 1)d\theta } \]
\[ = \left[ {\tan \theta - \theta } \right]_0^{{\pi \over 6}}\]
\[ = {1 \over {\sqrt 3 }} - {\pi \over 6}\]
Given, \[{k \over 6} = {1 \over {\sqrt 3 }} - {\pi \over 6} = {{2\sqrt 3 - \pi } \over 6}\]
\[ \Rightarrow k = 2\sqrt 3 - \pi \]