JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 5)
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 :
$$\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$$
is $${k \over 6}$$, then k is equal to :
$$2\sqrt 3 + \pi $$
$$3\sqrt 2 - \pi $$
$$3\sqrt 2 + \pi $$
$$2\sqrt 3 - \pi $$
Explanation
$$I = \int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{(1 - {x^2})}^{{3 \over 2}}}}}} dx$$
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, $${k \over 6} = {1 \over {\sqrt 3 }} - {\pi \over 6} = {{2\sqrt 3 - \pi } \over 6}$$
$$ \Rightarrow k = 2\sqrt 3 - \pi $$
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, $${k \over 6} = {1 \over {\sqrt 3 }} - {\pi \over 6} = {{2\sqrt 3 - \pi } \over 6}$$
$$ \Rightarrow k = 2\sqrt 3 - \pi $$
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