Integral de $$$\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)}\, dx$$$.
Solución
Reescribe el integrando:
$${\color{red}{\int{\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\cos^{3}{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}}$$
Multiplica el numerador y el denominador por un seno y expresa todo lo demás en términos del coseno, usando la fórmula $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ con $$$\alpha=2 x$$$:
$${\color{red}{\int{\frac{\cos^{3}{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(2 x \right)} \cos^{3}{\left(2 x \right)}}{1 - \cos^{2}{\left(2 x \right)}} d x}}}$$
Sea $$$u=\cos{\left(2 x \right)}$$$.
Entonces $$$du=\left(\cos{\left(2 x \right)}\right)^{\prime }dx = - 2 \sin{\left(2 x \right)} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\sin{\left(2 x \right)} dx = - \frac{du}{2}$$$.
La integral se convierte en
$${\color{red}{\int{\frac{\sin{\left(2 x \right)} \cos^{3}{\left(2 x \right)}}{1 - \cos^{2}{\left(2 x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{u^{3}}{2 \left(1 - u^{2}\right)}\right)d u}}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{u^{3}}{1 - u^{2}}$$$:
$${\color{red}{\int{\left(- \frac{u^{3}}{2 \left(1 - u^{2}\right)}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{u^{3}}{1 - u^{2}} d u}}{2}\right)}}$$
Como el grado del numerador no es menor que el grado del denominador, realiza la división larga de polinomios (los pasos pueden verse »):
$$- \frac{{\color{red}{\int{\frac{u^{3}}{1 - u^{2}} d u}}}}{2} = - \frac{{\color{red}{\int{\left(- u + \frac{u}{1 - u^{2}}\right)d u}}}}{2}$$
Integra término a término:
$$- \frac{{\color{red}{\int{\left(- u + \frac{u}{1 - u^{2}}\right)d u}}}}{2} = - \frac{{\color{red}{\left(- \int{u d u} + \int{\frac{u}{1 - u^{2}} d u}\right)}}}{2}$$
Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:
$$- \frac{\int{\frac{u}{1 - u^{2}} d u}}{2} + \frac{{\color{red}{\int{u d u}}}}{2}=- \frac{\int{\frac{u}{1 - u^{2}} d u}}{2} + \frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{2}=- \frac{\int{\frac{u}{1 - u^{2}} d u}}{2} + \frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{2}$$
Sea $$$v=1 - u^{2}$$$.
Entonces $$$dv=\left(1 - u^{2}\right)^{\prime }du = - 2 u du$$$ (los pasos pueden verse »), y obtenemos que $$$u du = - \frac{dv}{2}$$$.
La integral puede reescribirse como
$$\frac{u^{2}}{4} - \frac{{\color{red}{\int{\frac{u}{1 - u^{2}} d u}}}}{2} = \frac{u^{2}}{4} - \frac{{\color{red}{\int{\left(- \frac{1}{2 v}\right)d v}}}}{2}$$
Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=- \frac{1}{2}$$$ y $$$f{\left(v \right)} = \frac{1}{v}$$$:
$$\frac{u^{2}}{4} - \frac{{\color{red}{\int{\left(- \frac{1}{2 v}\right)d v}}}}{2} = \frac{u^{2}}{4} - \frac{{\color{red}{\left(- \frac{\int{\frac{1}{v} d v}}{2}\right)}}}{2}$$
La integral de $$$\frac{1}{v}$$$ es $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{u^{2}}{4} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = \frac{u^{2}}{4} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$
Recordemos que $$$v=1 - u^{2}$$$:
$$\frac{u^{2}}{4} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = \frac{u^{2}}{4} + \frac{\ln{\left(\left|{{\color{red}{\left(1 - u^{2}\right)}}}\right| \right)}}{4}$$
Recordemos que $$$u=\cos{\left(2 x \right)}$$$:
$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}^{2}}\right| \right)}}{4} + \frac{{\color{red}{u}}^{2}}{4} = \frac{\ln{\left(\left|{-1 + {\color{red}{\cos{\left(2 x \right)}}}^{2}}\right| \right)}}{4} + \frac{{\color{red}{\cos{\left(2 x \right)}}}^{2}}{4}$$
Por lo tanto,
$$\int{\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)} d x} = \frac{\ln{\left(\left|{\cos^{2}{\left(2 x \right)} - 1}\right| \right)}}{4} + \frac{\cos^{2}{\left(2 x \right)}}{4}$$
Añade la constante de integración:
$$\int{\sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)} d x} = \frac{\ln{\left(\left|{\cos^{2}{\left(2 x \right)} - 1}\right| \right)}}{4} + \frac{\cos^{2}{\left(2 x \right)}}{4}+C$$
Respuesta
$$$\int \sin^{2}{\left(2 x \right)} \cot^{3}{\left(2 x \right)}\, dx = \left(\frac{\ln\left(\left|{\cos^{2}{\left(2 x \right)} - 1}\right|\right)}{4} + \frac{\cos^{2}{\left(2 x \right)}}{4}\right) + C$$$A