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

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

Extraiga un factor de tangente y exprese todo lo demás en términos de la secante, usando la fórmula $$$\tan^2\left(x \right)=\sec^2\left(x \right)-1$$$:

$${\color{red}{\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}}$$

Sea $$$u=\sec{\left(x \right)}$$$.

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

La integral puede reescribirse como

$${\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

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

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

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

Recordemos que $$$u=\sec{\left(x \right)}$$$:

$$- {\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = - {\color{red}{\sec{\left(x \right)}}} + \frac{{\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$

Por lo tanto,

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

Añade la constante de integración:

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

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