Question

Evaluate : ${\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{x^8}{1-x^4}\times [{\sin }^{-1}(1-2x^2)+{\cos }^{-1}\left(2x\sqrt{1-x^2}\right)]dx$

• A. $\pi [\frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+{\tan }^{-1}\frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}]$
• B. $\pi [\frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+{\tan }^{-1}\frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}]$
• C. $\pi [\frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+{\tan }^{-1}\frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}]$
• D. $\pi [\frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+{\tan }^{-1}\frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}]$

Answer 1

The correct option is A $\pi [\frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+{\tan }^{-1}\frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}]$

In $N^r$ put $x=\sin \theta ,$ then ${\sin }^{-1}(1-2{\sin }^2\theta )+{\cos }^{-1}\left(2\sin \theta \cos \theta \right)={\sin }^{-1}\left(\cos 2\theta \right)+{\cos }^{-1}\left(\sin 2\theta \right)$ $=\frac{\pi }{2}-{\cos }^{-1}\left(\cos 2\theta \right)+\frac{\pi }{2}-{\sin }^{-1}\left(\sin 2\theta \right)$ $=\pi -2\theta -2\theta =\pi -4{\sin }^{-1}x$
$\therefore I={\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{x^8}{1-x^4}[\pi -4\sin -\mid x]dx$ $=2{\int }_0^{\frac{1}{\sqrt{2}}}\pi \frac{x^8}{1-x^4}dx+0$ by Prop.V $=2\pi {\int }_0^{\frac{1}{\sqrt{2}}}\frac{x^8-1+1}{1-x^4}dx$ $=2\pi {\int }_0^{\frac{1}{\sqrt{2}}}[-(x^4+1)+\frac{1}{(1-x^2)(1+x^2)}]dx$ $=2\pi {\int }_0^{\frac{1}{\sqrt{2}}}-(x^4+1)+\frac{1}{2}\{\frac{1}{1-x^2}+\frac{1}{1+x^2}\}dx$ $=2x{[-(\frac{x^5}{5}+x)+\frac{1}{4}\log \frac{1+x}{1-x}+\frac{1}{2}\tan -1x]}_0^{1/\sqrt{2}}$ $=\pi [\frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+{\tan }^{-1}\frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}]$
Ans: A