Finding the integral of rational function of sines $$ \int \frac{\sin x}{1+\sin x} \, \mathrm{d}x$$ How do I integrate this? I tried multiplying and dividing by $ (1- \sin x) $.

Using your idea, $\displaystyle\int\frac{\sin x}{1+\sin x}dx=\int\frac{\sin x}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}dx=\int\frac{\sin x-\sin^2 x}{\cos^2 x}dx$
$\displaystyle=\int\frac{\sin x}{\cos^2x}dx-\int\tan^2x \;dx=\int\sec x\tan xdx-\int(\sec^2x-1)dx=\sec x-\tan x+x+C$
