JEE Main 2020 Mathematics - Class 12 JEEM20Ma9520 Solution: If the value of the integral is , then k is equal to :

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

QJEEM20Ma9520: JEE Main 2020 Mathematics - Class 12 JEEM20Ma9520 Solution: If the value of the integral is , then k is equal to :