Integral de $$$\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)}$$$

Encontre $$$\int \tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)}\, dy$$$.

Elimine duas secantes e escreva o restante em termos da tangente, usando a fórmula $$$\sec^2\left( \alpha \right)=\tan^2\left( \alpha \right) + 1$$$ com $$$\alpha=y$$$:

$${\color{red}{\int{\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)} d y}}} = {\color{red}{\int{\left(\tan^{2}{\left(y \right)} + 1\right) \tan^{4}{\left(y \right)} \sec^{2}{\left(y \right)} d y}}}$$

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

Então $$$du=\left(\tan{\left(y \right)}\right)^{\prime }dy = \sec^{2}{\left(y \right)} dy$$$ (veja os passos »), e obtemos $$$\sec^{2}{\left(y \right)} dy = du$$$.

A integral torna-se

$${\color{red}{\int{\left(\tan^{2}{\left(y \right)} + 1\right) \tan^{4}{\left(y \right)} \sec^{2}{\left(y \right)} d y}}} = {\color{red}{\int{u^{4} \left(u^{2} + 1\right) d u}}}$$

Expand the expression:

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

Integre termo a termo:

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

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

$$\int{u^{6} d u} + {\color{red}{\int{u^{4} d u}}}=\int{u^{6} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=\int{u^{6} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=6$$$:

$$\frac{u^{5}}{5} + {\color{red}{\int{u^{6} d u}}}=\frac{u^{5}}{5} + {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=\frac{u^{5}}{5} + {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$

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

$$\frac{{\color{red}{u}}^{5}}{5} + \frac{{\color{red}{u}}^{7}}{7} = \frac{{\color{red}{\tan{\left(y \right)}}}^{5}}{5} + \frac{{\color{red}{\tan{\left(y \right)}}}^{7}}{7}$$

Portanto,

$$\int{\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)} d y} = \frac{\tan^{7}{\left(y \right)}}{7} + \frac{\tan^{5}{\left(y \right)}}{5}$$

Adicione a constante de integração:

$$\int{\tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)} d y} = \frac{\tan^{7}{\left(y \right)}}{7} + \frac{\tan^{5}{\left(y \right)}}{5}+C$$

$$$\int \tan^{4}{\left(y \right)} \sec^{4}{\left(y \right)}\, dy = \left(\frac{\tan^{7}{\left(y \right)}}{7} + \frac{\tan^{5}{\left(y \right)}}{5}\right) + C$$$A