# Integrating $\int{\frac{\sqrt{1-x^2}}{(x+\sqrt{1-x^2})^2} dx}$

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!

• I know you have a solution now, I added this part if you don't accept it please delete it Mar 16, 2014 at 20:29
• In ADDED there is a typo at the beginning since $\displaystyle{\cos^2(t) = {1 \color{#f00}{\LARGE+} \cos(2t) \over 2}}$ Mar 16, 2014 at 20:33
– user98602
Mar 16, 2014 at 20:35

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.
• 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 $u$. Mar 16, 2014 at 20:26