Integral de $$$\tan^{6}{\left(x \right)}$$$

Encontre $$$\int \tan^{6}{\left(x \right)}\, dx$$$.

Seja $$$u=\tan{\left(x \right)}$$$.

Então $$$x=\operatorname{atan}{\left(u \right)}$$$ e $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (as etapas podem ser vistas »).

Portanto,

$${\color{red}{\int{\tan^{6}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{u^{6}}{u^{2} + 1} d u}}}$$

Como o grau do numerador não é menor que o grau do denominador, realize a divisão longa de polinômios (os passos podem ser vistos »):

$${\color{red}{\int{\frac{u^{6}}{u^{2} + 1} d u}}} = {\color{red}{\int{\left(u^{4} - u^{2} + 1 - \frac{1}{u^{2} + 1}\right)d u}}}$$

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

$$- \int{u^{2} d u} + \int{u^{4} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{1 d u}}} = - \int{u^{2} d u} + \int{u^{4} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{u}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=4$$$:

$$u - \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{u^{4} d u}}}=u - \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=u - \int{u^{2} d u} - \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=2$$$:

$$\frac{u^{5}}{5} + u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{u^{2} d u}}}=\frac{u^{5}}{5} + u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=\frac{u^{5}}{5} + u - \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

A integral de $$$\frac{1}{u^{2} + 1}$$$ é $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$\frac{u^{5}}{5} - \frac{u^{3}}{3} + u - {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = \frac{u^{5}}{5} - \frac{u^{3}}{3} + u - {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

Recorde que $$$u=\tan{\left(x \right)}$$$:

$$- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}} - \frac{{\color{red}{u}}^{3}}{3} + \frac{{\color{red}{u}}^{5}}{5} = - \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} + {\color{red}{\tan{\left(x \right)}}} - \frac{{\color{red}{\tan{\left(x \right)}}}^{3}}{3} + \frac{{\color{red}{\tan{\left(x \right)}}}^{5}}{5}$$

Portanto,

$$\int{\tan^{6}{\left(x \right)} d x} = \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)} - \operatorname{atan}{\left(\tan{\left(x \right)} \right)}$$

Simplifique:

$$\int{\tan^{6}{\left(x \right)} d x} = - x + \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$$

Adicione a constante de integração:

$$\int{\tan^{6}{\left(x \right)} d x} = - x + \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+C$$

$$$\int \tan^{6}{\left(x \right)}\, dx = \left(- x + \frac{\tan^{5}{\left(x \right)}}{5} - \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}\right) + C$$$A