Integral de $$$\frac{1}{\sin{\left(x \right)}}$$$

Halla $$$\int \frac{1}{\sin{\left(x \right)}}\, dx$$$.

Reescribe el seno utilizando la fórmula del ángulo doble $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

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

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

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

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

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

Por lo tanto,

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

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

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

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

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

Por lo tanto,

$$\int{\frac{1}{\sin{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{1}{\sin{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}+C$$

$$$\int \frac{1}{\sin{\left(x \right)}}\, dx = \ln\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right|\right) + C$$$A