Integral de $$$\tan^{3}{\left(7 x \right)}$$$

Halla $$$\int \tan^{3}{\left(7 x \right)}\, dx$$$.

Sea $$$u=7 x$$$.

Entonces $$$du=\left(7 x\right)^{\prime }dx = 7 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{7}$$$.

Entonces,

$${\color{red}{\int{\tan^{3}{\left(7 x \right)} d x}}} = {\color{red}{\int{\frac{\tan^{3}{\left(u \right)}}{7} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{7}$$$ y $$$f{\left(u \right)} = \tan^{3}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\tan^{3}{\left(u \right)}}{7} d u}}} = {\color{red}{\left(\frac{\int{\tan^{3}{\left(u \right)} d u}}{7}\right)}}$$

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

Entonces $$$u=\operatorname{atan}{\left(v \right)}$$$ y $$$du=\left(\operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{dv}{v^{2} + 1}$$$ (los pasos se pueden ver »).

Por lo tanto,

$$\frac{{\color{red}{\int{\tan^{3}{\left(u \right)} d u}}}}{7} = \frac{{\color{red}{\int{\frac{v^{3}}{v^{2} + 1} d v}}}}{7}$$

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

$$\frac{{\color{red}{\int{\frac{v^{3}}{v^{2} + 1} d v}}}}{7} = \frac{{\color{red}{\int{\left(v - \frac{v}{v^{2} + 1}\right)d v}}}}{7}$$

Integra término a término:

$$\frac{{\color{red}{\int{\left(v - \frac{v}{v^{2} + 1}\right)d v}}}}{7} = \frac{{\color{red}{\left(\int{v d v} - \int{\frac{v}{v^{2} + 1} d v}\right)}}}{7}$$

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

$$- \frac{\int{\frac{v}{v^{2} + 1} d v}}{7} + \frac{{\color{red}{\int{v d v}}}}{7}=- \frac{\int{\frac{v}{v^{2} + 1} d v}}{7} + \frac{{\color{red}{\frac{v^{1 + 1}}{1 + 1}}}}{7}=- \frac{\int{\frac{v}{v^{2} + 1} d v}}{7} + \frac{{\color{red}{\left(\frac{v^{2}}{2}\right)}}}{7}$$

Sea $$$w=v^{2} + 1$$$.

Entonces $$$dw=\left(v^{2} + 1\right)^{\prime }dv = 2 v dv$$$ (los pasos pueden verse »), y obtenemos que $$$v dv = \frac{dw}{2}$$$.

Por lo tanto,

$$\frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{v}{v^{2} + 1} d v}}}}{7} = \frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{1}{2 w} d w}}}}{7}$$

Aplica la regla del factor constante $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(w \right)} = \frac{1}{w}$$$:

$$\frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{1}{2 w} d w}}}}{7} = \frac{v^{2}}{14} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{w} d w}}{2}\right)}}}{7}$$

La integral de $$$\frac{1}{w}$$$ es $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$:

$$\frac{v^{2}}{14} - \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{14} = \frac{v^{2}}{14} - \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{14}$$

Recordemos que $$$w=v^{2} + 1$$$:

$$\frac{v^{2}}{14} - \frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{14} = \frac{v^{2}}{14} - \frac{\ln{\left(\left|{{\color{red}{\left(v^{2} + 1\right)}}}\right| \right)}}{14}$$

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

$$- \frac{\ln{\left(1 + {\color{red}{v}}^{2} \right)}}{14} + \frac{{\color{red}{v}}^{2}}{14} = - \frac{\ln{\left(1 + {\color{red}{\tan{\left(u \right)}}}^{2} \right)}}{14} + \frac{{\color{red}{\tan{\left(u \right)}}}^{2}}{14}$$

Recordemos que $$$u=7 x$$$:

$$- \frac{\ln{\left(1 + \tan^{2}{\left({\color{red}{u}} \right)} \right)}}{14} + \frac{\tan^{2}{\left({\color{red}{u}} \right)}}{14} = - \frac{\ln{\left(1 + \tan^{2}{\left({\color{red}{\left(7 x\right)}} \right)} \right)}}{14} + \frac{\tan^{2}{\left({\color{red}{\left(7 x\right)}} \right)}}{14}$$

Por lo tanto,

$$\int{\tan^{3}{\left(7 x \right)} d x} = - \frac{\ln{\left(\tan^{2}{\left(7 x \right)} + 1 \right)}}{14} + \frac{\tan^{2}{\left(7 x \right)}}{14}$$

Añade la constante de integración:

$$\int{\tan^{3}{\left(7 x \right)} d x} = - \frac{\ln{\left(\tan^{2}{\left(7 x \right)} + 1 \right)}}{14} + \frac{\tan^{2}{\left(7 x \right)}}{14}+C$$

$$$\int \tan^{3}{\left(7 x \right)}\, dx = \left(- \frac{\ln\left(\tan^{2}{\left(7 x \right)} + 1\right)}{14} + \frac{\tan^{2}{\left(7 x \right)}}{14}\right) + C$$$A