I've done a pretty similar one, but for this one I don't get what I should get. I post the integral and my steps : $\displaystyle \omega=\int_0^{\frac{\pi}{4}} \frac{dx}{1+\sin (x)}$.

Let $\displaystyle x=2\arctan (u)$, $\displaystyle dx= \frac{2}{1+u^2}du$.

$\displaystyle \Rightarrow \omega=\int \frac{\frac{2}{1+u^2}}{1+\sin (2\arctan(u))}du$. Let $\displaystyle \theta=\arctan (u)$. $\displaystyle \Rightarrow \sin (2\arctan (u))=\sin (2\theta)=2\cos (\theta)\sin (\theta)$. But as $\displaystyle \cos (\theta)=\frac{1}{\sqrt{1+u^2}}$ and $\displaystyle \sin (\theta)=\frac{u}{\sqrt{1+u^2}}$, we have that $\displaystyle \omega= \int \frac{\frac{2}{1+u^2}}{\frac{2u}{1+u^2}}du=\int \frac{du}{u}=\ln |u| +C$, backsub I get $\displaystyle \ln |\tan \left( \frac{x}{2}\right)|+C$. Putting the limits I finally get that $\displaystyle \omega$ is worth $\displaystyle \ln|\tan \left( \frac{\pi}{8}\right) |+C$ since $\displaystyle \tan(0)=0$. But the answer should be $\displaystyle 2-\sqrt 2$ according to Mathematica...

I didn't see my error even when typing all this in Latex. Note that for simplicity I didn't take in count the limits of the integral, so it's not mathematically correct to say that $\displaystyle \omega$ is worth all the steps I did before to get to the last value (from which it should be equal to, but it seems it isn't).