If the value of the integral
$$\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$$

is ${k \over 6}$, then k is equal to :

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