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

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

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

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

Reescribe la secante como $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:

$$2 {\color{red}{\int{\sec{\left(x \right)} d x}}} = 2 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}$$

Expresa el coseno en función del seno utilizando la fórmula $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ y luego expresa el seno utilizando la fórmula del ángulo doble $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

$$2 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} = 2 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}$$

Multiplica el numerador y el denominador por $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

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

Sea $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.

Entonces $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

La integral se convierte en

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

La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$2 {\color{red}{\int{\frac{1}{u} d u}}} = 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recordemos que $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 2 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}$$

Por lo tanto,

$$\int{2 \sec{\left(x \right)} d x} = 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}$$

Añade la constante de integración:

$$\int{2 \sec{\left(x \right)} d x} = 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C$$

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