Integral de $$$\tan{\left(x \right)} \sec^{p}{\left(x \right)}$$$ con respecto a $$$x$$$

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

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 $$$\tan{\left(x \right)} \sec{\left(x \right)} dx = du$$$.

Entonces,

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

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

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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