Integral de $$$\cot^{4}{\left(3 x \right)}$$$

Halla $$$\int \cot^{4}{\left(3 x \right)}\, dx$$$.

Sea $$$u=3 x$$$.

Entonces $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{3}$$$.

Por lo tanto,

$${\color{red}{\int{\cot^{4}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\cot^{4}{\left(u \right)}}{3} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{3}$$$ y $$$f{\left(u \right)} = \cot^{4}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\cot^{4}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\cot^{4}{\left(u \right)} d u}}{3}\right)}}$$

Sea $$$v=\cot{\left(u \right)}$$$.

Entonces $$$dv=\left(\cot{\left(u \right)}\right)^{\prime }du = - \csc^{2}{\left(u \right)} du$$$ (los pasos pueden verse »), y obtenemos que $$$\csc^{2}{\left(u \right)} du = - dv$$$.

Por lo tanto,

$$\frac{{\color{red}{\int{\cot^{4}{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\int{\left(- \frac{v^{4}}{v^{2} + 1}\right)d v}}}}{3}$$

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=-1$$$ y $$$f{\left(v \right)} = \frac{v^{4}}{v^{2} + 1}$$$:

$$\frac{{\color{red}{\int{\left(- \frac{v^{4}}{v^{2} + 1}\right)d v}}}}{3} = \frac{{\color{red}{\left(- \int{\frac{v^{4}}{v^{2} + 1} d v}\right)}}}{3}$$

Como el grado del numerador no es menor que el grado del denominador, realiza la división larga de polinomios (los pasos pueden verse »):

$$- \frac{{\color{red}{\int{\frac{v^{4}}{v^{2} + 1} d v}}}}{3} = - \frac{{\color{red}{\int{\left(v^{2} - 1 + \frac{1}{v^{2} + 1}\right)d v}}}}{3}$$

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dv = c v$$$ con $$$c=1$$$:

$$- \frac{\int{v^{2} d v}}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} + \frac{{\color{red}{\int{1 d v}}}}{3} = - \frac{\int{v^{2} d v}}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} + \frac{{\color{red}{v}}}{3}$$

Aplica la regla de la potencia $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:

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

La integral de $$$\frac{1}{v^{2} + 1}$$$ es $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

$$- \frac{v^{3}}{9} + \frac{v}{3} - \frac{{\color{red}{\int{\frac{1}{v^{2} + 1} d v}}}}{3} = - \frac{v^{3}}{9} + \frac{v}{3} - \frac{{\color{red}{\operatorname{atan}{\left(v \right)}}}}{3}$$

Recordemos que $$$v=\cot{\left(u \right)}$$$:

$$- \frac{\operatorname{atan}{\left({\color{red}{v}} \right)}}{3} + \frac{{\color{red}{v}}}{3} - \frac{{\color{red}{v}}^{3}}{9} = - \frac{\operatorname{atan}{\left({\color{red}{\cot{\left(u \right)}}} \right)}}{3} + \frac{{\color{red}{\cot{\left(u \right)}}}}{3} - \frac{{\color{red}{\cot{\left(u \right)}}}^{3}}{9}$$

Recordemos que $$$u=3 x$$$:

$$\frac{\cot{\left({\color{red}{u}} \right)}}{3} - \frac{\cot^{3}{\left({\color{red}{u}} \right)}}{9} - \frac{\operatorname{atan}{\left(\cot{\left({\color{red}{u}} \right)} \right)}}{3} = \frac{\cot{\left({\color{red}{\left(3 x\right)}} \right)}}{3} - \frac{\cot^{3}{\left({\color{red}{\left(3 x\right)}} \right)}}{9} - \frac{\operatorname{atan}{\left(\cot{\left({\color{red}{\left(3 x\right)}} \right)} \right)}}{3}$$

Por lo tanto,

$$\int{\cot^{4}{\left(3 x \right)} d x} = - \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}$$

Añade la constante de integración:

$$\int{\cot^{4}{\left(3 x \right)} d x} = - \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}+C$$

$$$\int \cot^{4}{\left(3 x \right)}\, dx = \left(- \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}\right) + C$$$A