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How do I go about integrating:

$$\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$$

The common trigonometric substitutions don't seem to work here.

I think it requires to take some power of $x$ outside the square root but I am not able to solve further.

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HINT....If you want a trig substitution that works, try $x^2=\cos 2\theta$

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  • $\begingroup$ Works very well !! $\endgroup$ – Sudhanshu Dec 30 '15 at 18:23
  • $\begingroup$ $$\int\frac{1}{x}\sqrt{\frac {1-x^2}{1+x^2}}dx= \int\frac{1}{x^2}\sqrt{\frac {1-\cos 2t}{1+\cos 2t}}(-\sin2t)dt$$ $$=\int\frac{1}{\cos2t}\cdot\tan t \cdot(-\sin2t)dt=-\int \tan 2t \cdot\tan t dt$$ Then what ? $\endgroup$ – Angelo Mark Dec 30 '15 at 18:30
  • $\begingroup$ write $\tan 2t\tan t$ in terms of sin and cos, and get $\frac{2\sin^2t}{\cos 2t}$ and then use $2\sin^2t=1-\cos 2t$'and then you're pretty much home and dry. $\endgroup$ – David Quinn Dec 30 '15 at 19:23
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The answer is (after a request from OP) updated with more details

I suggest you to set $$ u=\sqrt{\frac{1-x^2}{1+x^2}}. $$ Then $$ x^2=\frac{1-u^2}{1+u^2} $$ and so $$ 2x\,dx=-\frac{4u}{(1+u^2)^2}\,du. $$ Thus $$ \begin{aligned} \int \frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,dx&=\int\frac{1}{2x^2}\sqrt{\frac{1-x^2}{1+x^2}}\,2x\,dx\\ &=\int\frac{1}{2}\frac{1+u^2}{1-u^2}u\cdot\Bigl(-\frac{4u}{(1+u^2)^2}\Bigr)\,du\\ &=-\int\frac{2u^2}{(1-u^2)(1+u^2)}\,du. \end{aligned} $$ Next, do a partial fraction decomposition, and you will end up with $$ \int\frac{1}{1+u^2}\,du-\int\frac{1}{1-u^2}\,du. $$ I guess you can take it from here? If not, ask for more details.

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  • $\begingroup$ I can do the latter .. However how did u simplified it ?? Because I am trying the same using that substitution and it looks messy.. $\endgroup$ – Sudhanshu Dec 30 '15 at 18:17
  • $\begingroup$ Got it Thanks.. $\endgroup$ – Sudhanshu Dec 30 '15 at 23:51

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