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