Integral de $$$\sin{\left(x \right)} \cos^{2}{\left(x \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$$.
Solución
Sea $$$u=\cos{\left(x \right)}$$$.
Entonces $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\sin{\left(x \right)} dx = - du$$$.
Entonces,
$${\color{red}{\int{\sin{\left(x \right)} \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- u^{2}\right)d u}}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = u^{2}$$$:
$${\color{red}{\int{\left(- u^{2}\right)d u}}} = {\color{red}{\left(- \int{u^{2} d u}\right)}}$$
Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:
$$- {\color{red}{\int{u^{2} d u}}}=- {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
Recordemos que $$$u=\cos{\left(x \right)}$$$:
$$- \frac{{\color{red}{u}}^{3}}{3} = - \frac{{\color{red}{\cos{\left(x \right)}}}^{3}}{3}$$
Por lo tanto,
$$\int{\sin{\left(x \right)} \cos^{2}{\left(x \right)} d x} = - \frac{\cos^{3}{\left(x \right)}}{3}$$
Añade la constante de integración:
$$\int{\sin{\left(x \right)} \cos^{2}{\left(x \right)} d x} = - \frac{\cos^{3}{\left(x \right)}}{3}+C$$
Respuesta
$$$\int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = - \frac{\cos^{3}{\left(x \right)}}{3} + C$$$A