Integral de $$$\frac{1}{2} - \frac{\cos{\left(6 t \right)}}{2}$$$

Encontre $$$\int \left(\frac{1}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)\, dt$$$.

Integre termo a termo:

$${\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)d t}}} = {\color{red}{\left(\int{\frac{1}{2} d t} - \int{\frac{\cos{\left(6 t \right)}}{2} d t}\right)}}$$

Aplique a regra da constante $$$\int c\, dt = c t$$$ usando $$$c=\frac{1}{2}$$$:

$$- \int{\frac{\cos{\left(6 t \right)}}{2} d t} + {\color{red}{\int{\frac{1}{2} d t}}} = - \int{\frac{\cos{\left(6 t \right)}}{2} d t} + {\color{red}{\left(\frac{t}{2}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(t \right)} = \cos{\left(6 t \right)}$$$:

$$\frac{t}{2} - {\color{red}{\int{\frac{\cos{\left(6 t \right)}}{2} d t}}} = \frac{t}{2} - {\color{red}{\left(\frac{\int{\cos{\left(6 t \right)} d t}}{2}\right)}}$$

Seja $$$u=6 t$$$.

Então $$$du=\left(6 t\right)^{\prime }dt = 6 dt$$$ (veja os passos »), e obtemos $$$dt = \frac{du}{6}$$$.

Assim,

$$\frac{t}{2} - \frac{{\color{red}{\int{\cos{\left(6 t \right)} d t}}}}{2} = \frac{t}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{2}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{6}$$$ e $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{t}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{2} = \frac{t}{2} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{2}$$

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

$$\frac{t}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{12} = \frac{t}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{12}$$

Recorde que $$$u=6 t$$$:

$$\frac{t}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{12} = \frac{t}{2} - \frac{\sin{\left({\color{red}{\left(6 t\right)}} \right)}}{12}$$

Portanto,

$$\int{\left(\frac{1}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)d t} = \frac{t}{2} - \frac{\sin{\left(6 t \right)}}{12}$$

Adicione a constante de integração:

$$\int{\left(\frac{1}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)d t} = \frac{t}{2} - \frac{\sin{\left(6 t \right)}}{12}+C$$

$$$\int \left(\frac{1}{2} - \frac{\cos{\left(6 t \right)}}{2}\right)\, dt = \left(\frac{t}{2} - \frac{\sin{\left(6 t \right)}}{12}\right) + C$$$A