Question

If ${\int }_{\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{x^4}{1-x^4}{\cos }^{-1}\frac{2x}{1+x^2}dx=k{\int }_0^{\frac{1}{\sqrt{3}}}\frac{x^4}{1-x^4}dx$,

then ${}^{\prime }{k}^{\prime }$ is equal to?

• A. $\pi$
• B. $2\pi$
• C. $2$
• D. $1$

Answer 1

Answer: (A)

$\pi$

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 odd function.
$\therefore I=\pi {\int }_0^{\frac{1}{\sqrt{3}}}\frac{x^4}{1-x^4}dx\Rightarrow k=\pi$