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

Question

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

$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}=\int\frac{\sin x}{2-\cos^{2}x}\mathrm {dx}$$

$u=\cos x$

$du=-\sin x dx$

$$-\int\frac{\mathrm {du}}{2-u^{2}}$$

¿Cómo podría continuar?

Answer 1

Una pista:

Escribe $$\frac{-1}{2-u^2}=\frac{-1}{(\sqrt{2}+u)(\sqrt{2}-u)}=\frac{-1}{2\sqrt{2}}\left(\frac{1}{\sqrt{2}+u}+\frac{1}{\sqrt{2}-u}\right)$$

Answer 2

$$-\int\frac{du}{2-u^{2}}\\ =-\frac 1 2 \int\frac{du}{1-u^2/2}$$

Set $t=u/\sqrt 2$ y $dt=\frac{du}{\sqrt 2}$

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

Answer 3

Observa, factoriza y utiliza las fracciones parciales de la siguiente manera $$-\int \frac{du}{2-u^2}=-\int \frac{du}{(\sqrt2-u)(\sqrt2+u)}$$ $$=-\frac{1}{2\sqrt 2}\int \left(\frac{1}{\sqrt 2-u}+\frac{1}{\sqrt 2+u}\right)\ du$$ $$=-\frac{1}{2\sqrt 2} \left(-\ln|\sqrt 2-u|+\ln|\sqrt2+u|\right)+C$$ $$=-\frac{1}{2\sqrt 2}\ln\left|\frac{\sqrt 2+u}{\sqrt 2-u}\right|+C$$ $$=\frac{1}{2\sqrt 2}\ln\left|\frac{\sqrt 2-u}{\sqrt 2+u}\right|+C$$ $$=\color{red}{\frac{1}{2\sqrt 2}\ln\left|\frac{\sqrt 2-\cos x}{\sqrt 2+\cos x}\right|+C}$$