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

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{\pi}{2}$$$ e $$$f{\left(x \right)} = \frac{1}{x^{2} \sqrt{x^{2} - 1}}$$$:

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

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

Então $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (os passos podem ser vistos »).

Além disso, segue-se que $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Portanto,

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

Use a identidade $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

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

Supondo que $$$\sinh{\left( u \right)} \ge 0$$$, obtemos o seguinte:

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

A integral pode ser reescrita como

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

Reescreva o integrando em termos da secante hiperbólica.:

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

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

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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