Find $$\int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx$$
What I have tried
First method was to try $u$ substitution
Let $u=\cos(x)$ then $-du=\sin(x)dx$ then $\sin(x)=\sqrt{1-u^2} $ which transforms our integral into
$$ \int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx=-\int \frac{1}{1+u\sqrt{1-u^2}}du=-\int \frac{1-u\sqrt{1-u^2}}{u^4-u^2+1}du$$
I think the might the denominator could be seperated since we get $u^2= \frac{1\pm \sqrt{3}i}{2}$ but I wouldn't know how to proceed after that.
Another method I have tried is this
Using the Weierstrass substitution
Let $\tan(x/2)=t$ , so $\sin(t) = \frac{2t}{1+t^2}$ , $\cos(t)=\frac{1-t^2}{1+t^2}$ and $dx=\frac{2dt}{1+t^2}$
Which transforms our integral as such
$$ \int \frac{\sin(x)}{1+\sin(x)\cos(x)}dx= \frac{4t}{t^4-2t^3+2t^2+2t+1} dt$$
I can't seem to find any roots by rational root theorem for the denonminator so I don't know how to proceed once again..
I wanted to post this as a comment, but I don't have enough points.
Anyway, I guess the trigonometric identity $\sin 2x = 2 \sin x \cdot \cos x$ may be useful for evaluating this integral. Good luck.