Integral de $$$\frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}}$$$

Encontre $$$\int \frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}}\, dx$$$.

Multiplique o numerador e o denominador por um seno e escreva todo o restante em termos do cosseno, usando a fórmula $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ com $$$\alpha=x$$$:

$${\color{red}{\int{\frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(x \right)} \cos^{3}{\left(x \right)}}{1 - \cos^{2}{\left(x \right)}} d x}}}$$

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

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

A integral torna-se

$${\color{red}{\int{\frac{\sin{\left(x \right)} \cos^{3}{\left(x \right)}}{1 - \cos^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{u^{3}}{1 - u^{2}}\right)d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=-1$$$ e $$$f{\left(u \right)} = \frac{u^{3}}{1 - u^{2}}$$$:

$${\color{red}{\int{\left(- \frac{u^{3}}{1 - u^{2}}\right)d u}}} = {\color{red}{\left(- \int{\frac{u^{3}}{1 - u^{2}} d u}\right)}}$$

Como o grau do numerador não é menor que o grau do denominador, realize a divisão longa de polinômios (os passos podem ser vistos »):

$$- {\color{red}{\int{\frac{u^{3}}{1 - u^{2}} d u}}} = - {\color{red}{\int{\left(- u + \frac{u}{1 - u^{2}}\right)d u}}}$$

Integre termo a termo:

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

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

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

Seja $$$v=1 - u^{2}$$$.

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

A integral torna-se

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

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

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

A integral de $$$\frac{1}{v}$$$ é $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Recorde que $$$v=1 - u^{2}$$$:

$$\frac{u^{2}}{2} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{u^{2}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(1 - u^{2}\right)}}}\right| \right)}}{2}$$

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

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

Portanto,

$$\int{\frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}} d x} = \frac{\ln{\left(\left|{\cos^{2}{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}$$

Adicione a constante de integração:

$$\int{\frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}} d x} = \frac{\ln{\left(\left|{\cos^{2}{\left(x \right)} - 1}\right| \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}+C$$

$$$\int \frac{\cos^{3}{\left(x \right)}}{\sin{\left(x \right)}}\, dx = \left(\frac{\ln\left(\left|{\cos^{2}{\left(x \right)} - 1}\right|\right)}{2} + \frac{\cos^{2}{\left(x \right)}}{2}\right) + C$$$A