integrate $\int\frac{\sin x}{1+\sin^{2}x}dx$ 
$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}$$

$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}=\int\frac{\sin x}{2-\cos^{2}x}\mathrm {dx}$$
$u=\cos x$
$du=-\sin x dx$
$$-\int\frac{\mathrm {du}}{2-u^{2}}$$
How could I continue?

$$-\int\frac{du}{2-u^{2}}\\
=-\frac 1 2 \int\frac{du}{1-u^2/2}$$
Set $t=u/\sqrt 2$ and $dt=\frac{du}{\sqrt 2}$
$$=-\frac{1}{\sqrt 2}\int \frac {1}{1-t^2}dt\\
=-\frac{\arctan\text{h}(t)}{\sqrt 2}+C\\
=-\frac{\arctan\text{h}(u/\sqrt 2)}{\sqrt 2}+C$$
