Integral de $$$\frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{40}$$$

Encontre $$$\int \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{40}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{\pi}{40}$$$ e $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{40} d x}}} = {\color{red}{\left(\frac{\pi \int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}{40}\right)}}$$

Seja $$$u=\sin{\left(x \right)}$$$.

Então $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (veja os passos »), e obtemos $$$\cos{\left(x \right)} dx = du$$$.

Assim,

$$\frac{\pi {\color{red}{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}}}{40} = \frac{\pi {\color{red}{\int{u d u}}}}{40}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:

$$\frac{\pi {\color{red}{\int{u d u}}}}{40}=\frac{\pi {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{40}=\frac{\pi {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{40}$$

Recorde que $$$u=\sin{\left(x \right)}$$$:

$$\frac{\pi {\color{red}{u}}^{2}}{80} = \frac{\pi {\color{red}{\sin{\left(x \right)}}}^{2}}{80}$$

Portanto,

$$\int{\frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{40} d x} = \frac{\pi \sin^{2}{\left(x \right)}}{80}$$

Adicione a constante de integração:

$$\int{\frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{40} d x} = \frac{\pi \sin^{2}{\left(x \right)}}{80}+C$$

$$$\int \frac{\pi \sin{\left(x \right)} \cos{\left(x \right)}}{40}\, dx = \frac{\pi \sin^{2}{\left(x \right)}}{80} + C$$$A