Question

${\int }_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}co{s}^{-1}\frac{2x}{1+x^2}dx=$

• A. $\frac{\pi }{2}\left[\frac{1}{2}\log (2+\sqrt{3})+\frac{\pi }{6}-\frac{2}{\sqrt{3}}\right]$
• B. $\frac{\pi }{4}\left[\frac{1}{2}\log (2+\sqrt{3})+\frac{\pi }{4}-\frac{2}{\sqrt{2}}\right]$
• C. $\frac{\pi }{2}\left[\frac{1}{2}\log (2-\sqrt{3})-\frac{\pi }{6}+\frac{2}{\sqrt{3}}\right]$
• D. $\frac{\pi }{2}\left[\frac{1}{3}\log (3+\sqrt{3})+\frac{\pi }{5}-\frac{1}{\sqrt{3}}\right]$

Answer 1

The correct option is A $\frac{\pi }{2}\left[\frac{1}{2}\log (2+\sqrt{3})+\frac{\pi }{6}-\frac{2}{\sqrt{3}}\right]$

Since ${\cos }^{-1}y=\frac{\pi }{2}-{\sin }^{-1}y$
$\therefore {\cos }^{-1}\left(\frac{2x}{1+x^2}\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\frac{\pi }{2}-2{\tan }^{-1}x$

$\therefore I={\int }_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}[\frac{\pi }{2}.\frac{x^4}{1-x^4}-\frac{x^4}{1+x^4}2{\tan }^{-1}x]dx$

[$\because \frac{x^4}{1-x^2}2{\tan }^{-1}x$ is an odd function]
$\therefore I=2\frac{\pi }{2}{\int }_0^{\frac{1}{\sqrt{3}}}(-1+\frac{1}{1-x^4})dx=\frac{\pi }{2}{\int }_0^{\frac{1}{\sqrt{3}}}(-2+\frac{1}{1-x^2}+\frac{1}{1+x^2})dx$
$=\frac{\pi }{2}{[-2x+\frac{1}{2.1}\log \frac{1+x}{1-x}+{\tan }^{-1}x]}_0^{\frac{1}{\sqrt{3}}}$
$=\frac{\pi }{2}[\frac{-2}{\sqrt{3}}+\frac{1}{2}\log \frac{\sqrt{3}+1}{\sqrt{3}-1}+\frac{\pi }{6}]$

$=\frac{\pi }{2}\left[\frac{1}{2}\log (2+\sqrt{3})+\frac{\pi }{6}-\frac{2}{\sqrt{3}}\right]$