Integral de $$$\sec^{4}{\left(\frac{x}{2} \right)}$$$

Halla $$$\int \sec^{4}{\left(\frac{x}{2} \right)}\, dx$$$.

Sea $$$u=\frac{x}{2}$$$.

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

Por lo tanto,

$${\color{red}{\int{\sec^{4}{\left(\frac{x}{2} \right)} d x}}} = {\color{red}{\int{2 \sec^{4}{\left(u \right)} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ y $$$f{\left(u \right)} = \sec^{4}{\left(u \right)}$$$:

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

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 $$$\alpha= u $$$:

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

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

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

La integral puede reescribirse como

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dv = c v$$$ con $$$c=1$$$:

$$2 \int{v^{2} d v} + 2 {\color{red}{\int{1 d v}}} = 2 \int{v^{2} d v} + 2 {\color{red}{v}}$$

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

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

Recordemos que $$$v=\tan{\left(u \right)}$$$:

$$2 {\color{red}{v}} + \frac{2 {\color{red}{v}}^{3}}{3} = 2 {\color{red}{\tan{\left(u \right)}}} + \frac{2 {\color{red}{\tan{\left(u \right)}}}^{3}}{3}$$

Recordemos que $$$u=\frac{x}{2}$$$:

$$2 \tan{\left({\color{red}{u}} \right)} + \frac{2 \tan^{3}{\left({\color{red}{u}} \right)}}{3} = 2 \tan{\left({\color{red}{\left(\frac{x}{2}\right)}} \right)} + \frac{2 \tan^{3}{\left({\color{red}{\left(\frac{x}{2}\right)}} \right)}}{3}$$

Por lo tanto,

$$\int{\sec^{4}{\left(\frac{x}{2} \right)} d x} = \frac{2 \tan^{3}{\left(\frac{x}{2} \right)}}{3} + 2 \tan{\left(\frac{x}{2} \right)}$$

Simplificar:

$$\int{\sec^{4}{\left(\frac{x}{2} \right)} d x} = \frac{2 \left(\tan^{2}{\left(\frac{x}{2} \right)} + 3\right) \tan{\left(\frac{x}{2} \right)}}{3}$$

Añade la constante de integración:

$$\int{\sec^{4}{\left(\frac{x}{2} \right)} d x} = \frac{2 \left(\tan^{2}{\left(\frac{x}{2} \right)} + 3\right) \tan{\left(\frac{x}{2} \right)}}{3}+C$$

$$$\int \sec^{4}{\left(\frac{x}{2} \right)}\, dx = \frac{2 \left(\tan^{2}{\left(\frac{x}{2} \right)} + 3\right) \tan{\left(\frac{x}{2} \right)}}{3} + C$$$A