Integral de $$$x \sqrt{1 - x^{2}}$$$

Halla $$$\int x \sqrt{1 - x^{2}}\, dx$$$.

Sea $$$u=1 - x^{2}$$$.

Entonces $$$du=\left(1 - x^{2}\right)^{\prime }dx = - 2 x dx$$$ (los pasos pueden verse »), y obtenemos que $$$x dx = - \frac{du}{2}$$$.

Por lo tanto,

$${\color{red}{\int{x \sqrt{1 - x^{2}} d x}}} = {\color{red}{\int{\left(- \frac{\sqrt{u}}{2}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{2}$$$ y $$$f{\left(u \right)} = \sqrt{u}$$$:

$${\color{red}{\int{\left(- \frac{\sqrt{u}}{2}\right)d u}}} = {\color{red}{\left(- \frac{\int{\sqrt{u} d u}}{2}\right)}}$$

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=\frac{1}{2}$$$:

$$- \frac{{\color{red}{\int{\sqrt{u} d u}}}}{2}=- \frac{{\color{red}{\int{u^{\frac{1}{2}} d u}}}}{2}=- \frac{{\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{2}=- \frac{{\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}}{2}$$

Recordemos que $$$u=1 - x^{2}$$$:

$$- \frac{{\color{red}{u}}^{\frac{3}{2}}}{3} = - \frac{{\color{red}{\left(1 - x^{2}\right)}}^{\frac{3}{2}}}{3}$$

Por lo tanto,

$$\int{x \sqrt{1 - x^{2}} d x} = - \frac{\left(1 - x^{2}\right)^{\frac{3}{2}}}{3}$$

Añade la constante de integración:

$$\int{x \sqrt{1 - x^{2}} d x} = - \frac{\left(1 - x^{2}\right)^{\frac{3}{2}}}{3}+C$$

$$$\int x \sqrt{1 - x^{2}}\, dx = - \frac{\left(1 - x^{2}\right)^{\frac{3}{2}}}{3} + C$$$A