Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 37540 by rahul 19 last updated on 14/Jun/18

$$\mathrm{For}x>1,$$ $$\int {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx=?$$

Answered by ajfour last updated on 14/Jun/18

$$letx=\tan \theta$$ $$x>1\Rightarrow n\pi +\frac{\pi }{4}<\theta <n\pi +\frac{\pi }{2}$$ $$\Rightarrow \frac{2x}{1+x^2}=\sin 2\theta$$ $$2n\pi +\frac{\pi }{2}<2\theta <2n\pi +\pi$$ $${\sin }^{-1}\left(\sin 2\theta \right)={\sin }^{-1}\sin (\pi -2\theta )$$ $$2m\pi <\pi -2\theta <2m\pi +\frac{\pi }{2}$$ $$m=0isappropriatesince$$ $$\pi -2\theta shouldliein[-\frac{\pi }{2},\frac{\pi }{2}]$$ $${\sin }^{-1}\sin (\pi -2\theta )=\pi -2\theta$$ $$=\pi -{2\tan }^{-1}x$$ $$I=\int (\pi -{2\tan }^{-1}x)dx$$ $$=\pi x-2(x{\tan }^{-1}x-\frac{1}{2}\int \frac{2xdx}{1+x^2})+c$$ $$I=\pi x-2x{\tan }^{-1}x+\ln \mid 1+x^2\mid +c.$$

Commented byrahul 19 last updated on 14/Jun/18

thank you sir. ����

Commented byrahul 19 last updated on 14/Jun/18

$$\mathrm{Ajfour}\mathrm{sir}\mathrm{pls}\mathrm{try}\mathrm{these}\mathrm{ques}...$$ $$\mathrm{Q}.\mathrm{no}.36700\mathrm{\&}36096.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com