Help with $\int \frac 1{\sqrt{a^2 - x^2}} \mathrm dx$

Question

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

In the second passage it become

$$\int \frac 1{a\sqrt{1 - \left(\frac xa\right)^2}}\,dx$$

So can someone explain me what kind operation is done at denominator?

Answer 1

$$\begin{align}\int \dfrac 1{\sqrt{a^2 - x^2}} \,\mathrm dx &= \dfrac 1{\sqrt{a^2\left(1 - \frac {x^2}{a^2}\right)}} \,\mathrm dx \\ \\ & = \int \dfrac 1{\sqrt {a^2}\cdot \sqrt{1 -\left(\frac xa\right)^2}} \,\mathrm dx \\ \\ & = \int \dfrac 1{|a|\sqrt{1 - \left(\frac xa\right)^2}}\,\mathrm dx\end{align}$$

If relevant to the transformation you are referring to, you might want to review the trigonometric substitution $(\text{see: }\;x = a\sin \theta)$ for how this progresses, if needed.