Question

If the value of the integral ${\int }_0^{\frac{1}{2}}\frac{x^2}{{(1-x^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx$ is $\frac{k}{6}$, then $k$ is equal to

• A. $2\sqrt{3}+\mathrm{\pi }$
• B. $3\sqrt{2}+\mathrm{\pi }$
• C. $3\sqrt{2}-\mathrm{\pi }$
• D. $2\sqrt{3}-\mathrm{\pi }$

Answer 1

The correct option is D

$2\sqrt{3}-\mathrm{\pi }$

Explanation for the correct option :

Evaluate the value of integral ,

Let us assume $x=\sin \theta$, then $dx=\cos \theta d\theta$.

Now,

$\begin{array}{rcl}{x}^2& =& {\left(\sin \theta \right)}^2\\ & =& {\sin }^2\theta \end{array}$

and

$\begin{array}{rcl}{(1-x^2)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}& =& {(1-{\sin }^2\theta )}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& {\left({\cos }^2\theta \right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ & =& {\left(\cos \theta \right)}^{2\times \frac{3}{2}}\\ & =& {\left(\cos \theta \right)}^3\\ & =& {\cos }^3\theta \end{array}$

The limits of integral will change as: for lower limit $x=0$ we have, $\theta =0$ and for the upper limit $x=\frac{1}{2}$ we have $\theta =\frac{\mathrm{\pi }}{6}$.

So substitute all these changes on the given integral ${\int }_0^{\frac{1}{2}}\frac{x^2}{{(1-x^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx$.

$\begin{array}{rcl}{\int }_0^{\frac{1}{2}}\frac{x^2}{{(1-x^2)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}dx& =& {\int }_0^{\frac{\mathrm{\pi }}{6}}\frac{{\sin }^2\theta }{{\cos }^3\theta }\cos \theta d\theta \\ & =& {\int }_0^{\frac{\mathrm{\pi }}{6}}\frac{{\sin }^2\theta }{{\cos }^2\theta }d\theta \\ & =& {\int }_0^{\frac{\mathrm{\pi }}{6}}{\tan }^2\theta d\theta \end{array}$

Now using the identity ${\sec }^2\theta -{\tan }^2\theta =1$ the integral can be evaluated as:

$\begin{array}{rcl}{\int }_0^{\frac{\mathrm{\pi }}{6}}{\tan }^2\theta d\theta & =& {\int }_0^{\frac{\mathrm{\pi }}{6}}\left({\sec }^2\theta -1\right)d\theta \\ & =& {\int }_0^{\frac{\mathrm{\pi }}{6}}{\sec }^2\theta d\theta -{\int }_0^{\frac{\mathrm{\pi }}{6}}1d\theta \\ & =& {\left[\tan \theta \right]}_0^{\frac{\mathrm{\pi }}{6}}-{\left[\theta \right]}_0^{\frac{\mathrm{\pi }}{6}}\\ & =& \tan \frac{\mathrm{\pi }}{6}-0-\left(\frac{\mathrm{\pi }}{6}-0\right)\\ & =& \frac{1}{\sqrt{3}}-\frac{\mathrm{\pi }}{6}\\ & =& \frac{\sqrt{3}}{3}-\frac{\mathrm{\pi }}{6}\\ & =& \frac{2\sqrt{3}}{6}-\frac{\mathrm{\pi }}{6}\\ & =& \frac{2\sqrt{3}-\mathrm{\pi }}{6}\end{array}$

Therefore, the value of integral was given as $\frac{k}{6}$ so the value of $k$ is $2\sqrt{3}-\mathrm{\pi }$.

Hence, the correct option is D.