Finding the integral of rational function of sines

Question

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

Answer 1

Using your idea, $\displaystyle\int\frac{\sin x}{1+\sin x}dx=\int\frac{\sin x}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}dx=\int\frac{\sin x-\sin^2 x}{\cos^2 x}dx$

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

Answer 2

First, let's simplify things just a bit and write

$$\frac{\sin x}{1+\sin x}=1-\frac{1}{1+\sin x}$$

Then, applying the Wierstrauss substitution $u=\tan(x/2)$ with $\sin x=\frac{2u}{1+u^2}$ and $dx=\frac{2du}{1+u^2}$ reveals that

$$\begin{align} \int\frac{\sin x}{1+\sin x}dx&=\int \left(1-\frac{1}{1+\sin x}\right)dx\\\\ &=x-\int\frac{2}{1+u^2+2u}du\\\\ &=x-2\int\frac{1}{(u+1)^2}\\\\ &=x+\frac{2}{1+\tan(x/2)}+C \end{align}$$

Thus, we have

$$\bbox[5px,border:2px solid #C0A000]{\int \frac{\sin x}{1+\sin x}dx=x+\frac{2}{1+\tan(x/2)}+C}$$