Integral de $$$\frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}}$$$

Halla $$$\int \frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}}\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{\pi}{2}$$$ y $$$f{\left(x \right)} = \frac{1}{x^{2} \sqrt{x^{2} - 1}}$$$:

$${\color{red}{\int{\frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}} d x}}} = {\color{red}{\left(\frac{\pi \int{\frac{1}{x^{2} \sqrt{x^{2} - 1}} d x}}{2}\right)}}$$

Sea $$$x=\cosh{\left(u \right)}$$$.

Entonces $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (los pasos pueden verse »).

Además, se sigue que $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Por lo tanto,

$$$\frac{1}{x^{2} \sqrt{x^{2} - 1}} = \frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1} \cosh^{2}{\left( u \right)}}$$$

Utiliza la identidad $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1} \cosh^{2}{\left( u \right)}}=\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}} \cosh^{2}{\left( u \right)}}$$$

Suponiendo que $$$\sinh{\left( u \right)} \ge 0$$$, obtenemos lo siguiente:

$$$\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}} \cosh^{2}{\left( u \right)}} = \frac{1}{\sinh{\left( u \right)} \cosh^{2}{\left( u \right)}}$$$

La integral se convierte en

$$\frac{\pi {\color{red}{\int{\frac{1}{x^{2} \sqrt{x^{2} - 1}} d x}}}}{2} = \frac{\pi {\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}}}{2}$$

Reescribe el integrando en términos de la secante hiperbólica:

$$\frac{\pi {\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}}}{2} = \frac{\pi {\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{2}$$

La integral de $$$\operatorname{sech}^{2}{\left(u \right)}$$$ es $$$\int{\operatorname{sech}^{2}{\left(u \right)} d u} = \tanh{\left(u \right)}$$$:

$$\frac{\pi {\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{2} = \frac{\pi {\color{red}{\tanh{\left(u \right)}}}}{2}$$

Recordemos que $$$u=\operatorname{acosh}{\left(x \right)}$$$:

$$\frac{\pi \tanh{\left({\color{red}{u}} \right)}}{2} = \frac{\pi \tanh{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}}{2}$$

Por lo tanto,

$$\int{\frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}} d x} = \frac{\pi \sqrt{x - 1} \sqrt{x + 1}}{2 x}$$

Añade la constante de integración:

$$\int{\frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}} d x} = \frac{\pi \sqrt{x - 1} \sqrt{x + 1}}{2 x}+C$$

$$$\int \frac{\pi}{2 x^{2} \sqrt{x^{2} - 1}}\, dx = \frac{\pi \sqrt{x - 1} \sqrt{x + 1}}{2 x} + C$$$A