Integral de $$$\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}$$$

Halla $$$\int \left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}\, dt$$$.

Sea $$$u=\cos{\left(t \right)} + 1$$$.

Entonces $$$du=\left(\cos{\left(t \right)} + 1\right)^{\prime }dt = - \sin{\left(t \right)} dt$$$ (los pasos pueden verse »), y obtenemos que $$$\sin{\left(t \right)} dt = - du$$$.

La integral puede reescribirse como

$${\color{red}{\int{\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)} d t}}} = {\color{red}{\int{\left(- u\right)d u}}}$$

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

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

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

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

Recordemos que $$$u=\cos{\left(t \right)} + 1$$$:

$$- \frac{{\color{red}{u}}^{2}}{2} = - \frac{{\color{red}{\left(\cos{\left(t \right)} + 1\right)}}^{2}}{2}$$

Por lo tanto,

$$\int{\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)} d t} = - \frac{\left(\cos{\left(t \right)} + 1\right)^{2}}{2}$$

Añade la constante de integración:

$$\int{\left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)} d t} = - \frac{\left(\cos{\left(t \right)} + 1\right)^{2}}{2}+C$$

$$$\int \left(\cos{\left(t \right)} + 1\right) \sin{\left(t \right)}\, dt = - \frac{\left(\cos{\left(t \right)} + 1\right)^{2}}{2} + C$$$A