Integral de $$$\frac{1}{\sin^{2}{\left(2 x \right)}}$$$

Halla $$$\int \frac{1}{\sin^{2}{\left(2 x \right)}}\, dx$$$.

Sea $$$u=2 x$$$.

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

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{1}{\sin^{2}{\left(u \right)}}$$$:

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

Reescribe el integrando en términos de la cosecante:

$$\frac{{\color{red}{\int{\frac{1}{\sin^{2}{\left(u \right)}} d u}}}}{2} = \frac{{\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}}{2}$$

La integral de $$$\csc^{2}{\left(u \right)}$$$ es $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\left(- \cot{\left(u \right)}\right)}}}{2}$$

Recordemos que $$$u=2 x$$$:

$$- \frac{\cot{\left({\color{red}{u}} \right)}}{2} = - \frac{\cot{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$

Por lo tanto,

$$\int{\frac{1}{\sin^{2}{\left(2 x \right)}} d x} = - \frac{\cot{\left(2 x \right)}}{2}$$

Añade la constante de integración:

$$\int{\frac{1}{\sin^{2}{\left(2 x \right)}} d x} = - \frac{\cot{\left(2 x \right)}}{2}+C$$

$$$\int \frac{1}{\sin^{2}{\left(2 x \right)}}\, dx = - \frac{\cot{\left(2 x \right)}}{2} + C$$$A