Integral de $$$\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)}$$$
Calculadora relacionada: Calculadora de Integrais Definidas e Impróprias
Sua entrada
Encontre $$$\int \sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)}\, dx$$$.
Solução
Reescreva o integrando:
$${\color{red}{\int{\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\cos^{3}{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}}$$
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 $$$\alpha=2 x$$$:
$${\color{red}{\int{\frac{\cos^{3}{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(2 x \right)} \cos^{3}{\left(2 x \right)}}{1 - \cos^{2}{\left(2 x \right)}} d x}}}$$
Seja $$$u=\cos{\left(2 x \right)}$$$.
Então $$$du=\left(\cos{\left(2 x \right)}\right)^{\prime }dx = - 2 \sin{\left(2 x \right)} dx$$$ (veja os passos »), e obtemos $$$\sin{\left(2 x \right)} dx = - \frac{du}{2}$$$.
Logo,
$${\color{red}{\int{\frac{\sin{\left(2 x \right)} \cos^{3}{\left(2 x \right)}}{1 - \cos^{2}{\left(2 x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{u^{3}}{2 \left(1 - u^{2}\right)}\right)d u}}}$$
Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=- \frac{1}{2}$$$ e $$$f{\left(u \right)} = \frac{u^{3}}{1 - u^{2}}$$$:
$${\color{red}{\int{\left(- \frac{u^{3}}{2 \left(1 - u^{2}\right)}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{u^{3}}{1 - u^{2}} d u}}{2}\right)}}$$
Como o grau do numerador não é menor que o grau do denominador, realize a divisão longa de polinômios (os passos podem ser vistos »):
$$- \frac{{\color{red}{\int{\frac{u^{3}}{1 - u^{2}} d u}}}}{2} = - \frac{{\color{red}{\int{\left(- u + \frac{u}{1 - u^{2}}\right)d u}}}}{2}$$
Integre termo a termo:
$$- \frac{{\color{red}{\int{\left(- u + \frac{u}{1 - u^{2}}\right)d u}}}}{2} = - \frac{{\color{red}{\left(- \int{u d u} + \int{\frac{u}{1 - u^{2}} d u}\right)}}}{2}$$
Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:
$$- \frac{\int{\frac{u}{1 - u^{2}} d u}}{2} + \frac{{\color{red}{\int{u d u}}}}{2}=- \frac{\int{\frac{u}{1 - u^{2}} d u}}{2} + \frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=- \frac{\int{\frac{u}{1 - u^{2}} d u}}{2} + \frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$
Seja $$$v=1 - u^{2}$$$.
Então $$$dv=\left(1 - u^{2}\right)^{\prime }du = - 2 u du$$$ (veja os passos »), e obtemos $$$u du = - \frac{dv}{2}$$$.
Logo,
$$\frac{u^{2}}{4} - \frac{{\color{red}{\int{\frac{u}{1 - u^{2}} d u}}}}{2} = \frac{u^{2}}{4} - \frac{{\color{red}{\int{\left(- \frac{1}{2 v}\right)d v}}}}{2}$$
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 $$$f{\left(v \right)} = \frac{1}{v}$$$:
$$\frac{u^{2}}{4} - \frac{{\color{red}{\int{\left(- \frac{1}{2 v}\right)d v}}}}{2} = \frac{u^{2}}{4} - \frac{{\color{red}{\left(- \frac{\int{\frac{1}{v} d v}}{2}\right)}}}{2}$$
A integral de $$$\frac{1}{v}$$$ é $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{u^{2}}{4} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = \frac{u^{2}}{4} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$
Recorde que $$$v=1 - u^{2}$$$:
$$\frac{u^{2}}{4} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = \frac{u^{2}}{4} + \frac{\ln{\left(\left|{{\color{red}{\left(1 - u^{2}\right)}}}\right| \right)}}{4}$$
Recorde que $$$u=\cos{\left(2 x \right)}$$$:
$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}^{2}}\right| \right)}}{4} + \frac{{\color{red}{u}}^{2}}{4} = \frac{\ln{\left(\left|{-1 + {\color{red}{\cos{\left(2 x \right)}}}^{2}}\right| \right)}}{4} + \frac{{\color{red}{\cos{\left(2 x \right)}}}^{2}}{4}$$
Portanto,
$$\int{\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)} d x} = \frac{\ln{\left(\left|{\cos^{2}{\left(2 x \right)} - 1}\right| \right)}}{4} + \frac{\cos^{2}{\left(2 x \right)}}{4}$$
Adicione a constante de integração:
$$\int{\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)} d x} = \frac{\ln{\left(\left|{\cos^{2}{\left(2 x \right)} - 1}\right| \right)}}{4} + \frac{\cos^{2}{\left(2 x \right)}}{4}+C$$
Resposta
$$$\int \sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)}\, dx = \left(\frac{\ln\left(\left|{\cos^{2}{\left(2 x \right)} - 1}\right|\right)}{4} + \frac{\cos^{2}{\left(2 x \right)}}{4}\right) + C$$$A