Integral de $$$\sin{\left(2 t \right)}$$$

Encontre $$$\int \sin{\left(2 t \right)}\, dt$$$.

Seja $$$u=2 t$$$.

Então $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (veja os passos »), e obtemos $$$dt = \frac{du}{2}$$$.

Logo,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

A integral do seno é $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Recorde que $$$u=2 t$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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