Integral de $$$\frac{1}{\sqrt{- a^{2} + x^{2}}}$$$ con respecto a $$$x$$$

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

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

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

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

Entonces,

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

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

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

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

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

Entonces,

$${\color{red}{\int{\frac{1}{\sqrt{- a^{2} + x^{2}}} d x}}} = {\color{red}{\int{1 d u}}}$$

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$

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

$${\color{red}{u}} = {\color{red}{\operatorname{acosh}{\left(\frac{x}{\left|{a}\right|} \right)}}}$$

Por lo tanto,

$$\int{\frac{1}{\sqrt{- a^{2} + x^{2}}} d x} = \operatorname{acosh}{\left(\frac{x}{\left|{a}\right|} \right)}$$

Añade la constante de integración:

$$\int{\frac{1}{\sqrt{- a^{2} + x^{2}}} d x} = \operatorname{acosh}{\left(\frac{x}{\left|{a}\right|} \right)}+C$$

$$$\int \frac{1}{\sqrt{- a^{2} + x^{2}}}\, dx = \operatorname{acosh}{\left(\frac{x}{\left|{a}\right|} \right)} + C$$$A