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

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

Para la integral $$$\int{\sec^{3}{\left(x \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\sec{\left(x \right)}$$$ y $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Entonces $$$\operatorname{du}=\left(\sec{\left(x \right)}\right)^{\prime }dx=\tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (los pasos pueden verse »).

Por lo tanto,

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

Aplica la fórmula $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:

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

Expandir:

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

La integral de una diferencia es la diferencia de las integrales:

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

Por lo tanto, obtenemos la siguiente ecuación lineal simple con respecto a la integral:

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

Al resolverlo, obtenemos que

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

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

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

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 $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

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

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

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

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 $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

Por lo tanto,

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

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

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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