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I am a little bit lost with integral: $$\int{\frac{\sqrt{1-x^2}}{(x+\sqrt{1-x^2})^2} dx}$$

I have already worked on in and done substitution $x = \sin(t)$:

This brings me to: $$\int{\frac{\cos(t)^2}{(\sin(t)+\cos(t))^2}dt}$$

Further treating denominator to achieve:

$$\int{\frac{\cos(t)^2}{\sin(2t)+1}dt}$$

I can split this fraction into two integrals by doing $\cos(t)^2 = 1-\sin(t)^2$ but this doesn't help me to solve the integral further.

Please, can you show me how to continue to "break it" :-)

Best thanks!

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  • $\begingroup$ I know you have a solution now, I added this part if you don't accept it please delete it $\endgroup$
    – Semsem
    Mar 16, 2014 at 20:29
  • $\begingroup$ In ADDED there is a typo at the beginning since $\displaystyle{\cos^2(t) = {1 \color{#f00}{\LARGE+} \cos(2t) \over 2}}$ $\endgroup$
    – Felix Marin
    Mar 16, 2014 at 20:33
  • $\begingroup$ @Semsem Please do not add your answers to the OP... submit it as an actual answer. $\endgroup$
    – user98602
    Mar 16, 2014 at 20:35

2 Answers 2

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We do a follow up to OP's substitution. We need to find $$\int \frac{\cos^2 t}{(\sin t+\cos t)^2}\,dt.$$

Note that $\sin t+\cos t=\sqrt{2}\cos(t-\pi/4)$. If we let $u=t-\pi/4$, then $\cos t=\cos(u+\pi/4)$. Expand. We get $\cos t=\frac{1}{\sqrt{2}}(\cos u-\sin u)$. Square. The rest is straightforward.

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  • $\begingroup$ Actually this is what I am looking for. Thank's for the kind advice. Regards! $\endgroup$
    – Jan Vorisek
    Mar 16, 2014 at 20:23
  • $\begingroup$ You are welcome. Complicating the numerator slightly is fine, as long as the denominator becomes nice. We could also have let $x+\sqrt{1-x^2}$ be $u$. $\endgroup$
    – André Nicolas
    Mar 16, 2014 at 20:26
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Instead of doing a trig substitution, expand the denominator and simplify the integrand.

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  • $\begingroup$ Thanks. I have done it, but i am not able to simplify it any better. Getting the same as with the substitution. $\endgroup$
    – Jan Vorisek
    Mar 16, 2014 at 20:14
  • $\begingroup$ Sorry I was mistaken- on mobile and did it in my head. $\endgroup$
    – Jason Zimba
    Mar 16, 2014 at 20:28

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