Integral de $$$\tan^{4}{\left(x \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \tan^{4}{\left(x \right)}\, dx$$$.
Solución
Sea $$$u=\tan{\left(x \right)}$$$.
Entonces $$$x=\operatorname{atan}{\left(u \right)}$$$ y $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (los pasos se pueden ver »).
La integral puede reescribirse como
$${\color{red}{\int{\tan^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{u^{4}}{u^{2} + 1} d u}}}$$
Como el grado del numerador no es menor que el grado del denominador, realiza la división larga de polinomios (los pasos pueden verse »):
$${\color{red}{\int{\frac{u^{4}}{u^{2} + 1} d u}}} = {\color{red}{\int{\left(u^{2} - 1 + \frac{1}{u^{2} + 1}\right)d u}}}$$
Integra término a término:
$${\color{red}{\int{\left(u^{2} - 1 + \frac{1}{u^{2} + 1}\right)d u}}} = {\color{red}{\left(- \int{1 d u} + \int{u^{2} d u} + \int{\frac{1}{u^{2} + 1} d u}\right)}}$$
Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:
$$\int{u^{2} d u} + \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{1 d u}}} = \int{u^{2} d u} + \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{u}}$$
Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:
$$- u + \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{u^{2} d u}}}=- u + \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- u + \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$
La integral de $$$\frac{1}{u^{2} + 1}$$$ es $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:
$$\frac{u^{3}}{3} - u + {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = \frac{u^{3}}{3} - u + {\color{red}{\operatorname{atan}{\left(u \right)}}}$$
Recordemos que $$$u=\tan{\left(x \right)}$$$:
$$\operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} - {\color{red}{\tan{\left(x \right)}}} + \frac{{\color{red}{\tan{\left(x \right)}}}^{3}}{3}$$
Por lo tanto,
$$\int{\tan^{4}{\left(x \right)} d x} = \frac{\tan^{3}{\left(x \right)}}{3} - \tan{\left(x \right)} + \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$
Simplificar:
$$\int{\tan^{4}{\left(x \right)} d x} = x + \frac{\tan^{3}{\left(x \right)}}{3} - \tan{\left(x \right)}$$
Añade la constante de integración:
$$\int{\tan^{4}{\left(x \right)} d x} = x + \frac{\tan^{3}{\left(x \right)}}{3} - \tan{\left(x \right)}+C$$
Respuesta
$$$\int \tan^{4}{\left(x \right)}\, dx = \left(x + \frac{\tan^{3}{\left(x \right)}}{3} - \tan{\left(x \right)}\right) + C$$$A