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

Halla $$$\int \sec^{4}{\left(x \right)}\, dx$$$.

Extrae dos secantes y escribe todo lo demás en términos de la tangente, utilizando la fórmula $$$\sec^2\left( \alpha \right)=\tan^2\left( \alpha \right) + 1$$$ con $$$\alpha=x$$$:

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

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

Entonces $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\sec^{2}{\left(x \right)} dx = du$$$.

Por lo tanto,

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

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

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

Por lo tanto,

$$\int{\sec^{4}{\left(x \right)} d x} = \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$$

Añade la constante de integración:

$$\int{\sec^{4}{\left(x \right)} d x} = \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+C$$

$$$\int \sec^{4}{\left(x \right)}\, dx = \left(\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}\right) + C$$$A