Integral de $$$\frac{1}{1 - \cos{\left(2 x \right)}}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \frac{1}{1 - \cos{\left(2 x \right)}}\, dx$$$.
Solución
Sea $$$u=2 x$$$.
Entonces $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{2}$$$.
La integral puede reescribirse como
$${\color{red}{\int{\frac{1}{1 - \cos{\left(2 x \right)}} d x}}} = {\color{red}{\int{\left(- \frac{1}{2 \left(\cos{\left(u \right)} - 1\right)}\right)d u}}}$$
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 $$$f{\left(u \right)} = \frac{1}{\cos{\left(u \right)} - 1}$$$:
$${\color{red}{\int{\left(- \frac{1}{2 \left(\cos{\left(u \right)} - 1\right)}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{\cos{\left(u \right)} - 1} d u}}{2}\right)}}$$
Reescribe el coseno usando la fórmula del ángulo doble $$$\cos\left( u \right)=1-2\sin^2\left(\frac{ u }{2}\right)$$$ y simplifica:
$$- \frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)} - 1} d u}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{1}{2 \sin^{2}{\left(\frac{u}{2} \right)}}\right)d u}}}}{2}$$
Sea $$$v=\frac{u}{2}$$$.
Entonces $$$dv=\left(\frac{u}{2}\right)^{\prime }du = \frac{du}{2}$$$ (los pasos pueden verse »), y obtenemos que $$$du = 2 dv$$$.
La integral se convierte en
$$- \frac{{\color{red}{\int{\left(- \frac{1}{2 \sin^{2}{\left(\frac{u}{2} \right)}}\right)d u}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{1}{\sin^{2}{\left(v \right)}}\right)d v}}}}{2}$$
Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=-1$$$ y $$$f{\left(v \right)} = \frac{1}{\sin^{2}{\left(v \right)}}$$$:
$$- \frac{{\color{red}{\int{\left(- \frac{1}{\sin^{2}{\left(v \right)}}\right)d v}}}}{2} = - \frac{{\color{red}{\left(- \int{\frac{1}{\sin^{2}{\left(v \right)}} d v}\right)}}}{2}$$
Reescribe el integrando en términos de la cosecante:
$$\frac{{\color{red}{\int{\frac{1}{\sin^{2}{\left(v \right)}} d v}}}}{2} = \frac{{\color{red}{\int{\csc^{2}{\left(v \right)} d v}}}}{2}$$
La integral de $$$\csc^{2}{\left(v \right)}$$$ es $$$\int{\csc^{2}{\left(v \right)} d v} = - \cot{\left(v \right)}$$$:
$$\frac{{\color{red}{\int{\csc^{2}{\left(v \right)} d v}}}}{2} = \frac{{\color{red}{\left(- \cot{\left(v \right)}\right)}}}{2}$$
Recordemos que $$$v=\frac{u}{2}$$$:
$$- \frac{\cot{\left({\color{red}{v}} \right)}}{2} = - \frac{\cot{\left({\color{red}{\left(\frac{u}{2}\right)}} \right)}}{2}$$
Recordemos que $$$u=2 x$$$:
$$- \frac{\cot{\left(\frac{{\color{red}{u}}}{2} \right)}}{2} = - \frac{\cot{\left(\frac{{\color{red}{\left(2 x\right)}}}{2} \right)}}{2}$$
Por lo tanto,
$$\int{\frac{1}{1 - \cos{\left(2 x \right)}} d x} = - \frac{\cot{\left(x \right)}}{2}$$
Añade la constante de integración:
$$\int{\frac{1}{1 - \cos{\left(2 x \right)}} d x} = - \frac{\cot{\left(x \right)}}{2}+C$$
Respuesta
$$$\int \frac{1}{1 - \cos{\left(2 x \right)}}\, dx = - \frac{\cot{\left(x \right)}}{2} + C$$$A