Integral de $$$\frac{\cos{\left(\frac{x}{2} - 1 \right)}}{\sin^{2}{\left(\frac{x}{2} - 1 \right)}}$$$

Encontre $$$\int \frac{\cos{\left(\frac{x}{2} - 1 \right)}}{\sin^{2}{\left(\frac{x}{2} - 1 \right)}}\, dx$$$.

Seja $$$u=\sin{\left(\frac{x}{2} - 1 \right)}$$$.

Então $$$du=\left(\sin{\left(\frac{x}{2} - 1 \right)}\right)^{\prime }dx = \frac{\cos{\left(\frac{x}{2} - 1 \right)}}{2} dx$$$ (veja os passos »), e obtemos $$$\cos{\left(\frac{x}{2} - 1 \right)} dx = 2 du$$$.

A integral torna-se

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

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

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

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-2$$$:

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

Recorde que $$$u=\sin{\left(\frac{x}{2} - 1 \right)}$$$:

$$- 2 {\color{red}{u}}^{-1} = - 2 {\color{red}{\sin{\left(\frac{x}{2} - 1 \right)}}}^{-1}$$

Portanto,

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

Adicione a constante de integração:

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

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