1/sq root of (5-x^2) and how do you get there??
integrate 1/sqrt(5 - x^2) - Wolfram|Alpha ... click on 'show steps'.
Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
Well I showed you the very first step. The next is to bring the 5 outside of the square root...
$\displaystyle \frac{1}{\sqrt{5}\sqrt{(1 - \frac{1}{5}x^2)}}$
Then bring the fifth inside of the square, with x, and bring the 5 in the denominator into its own fraction...
$\displaystyle \frac{1}{\sqrt{5}}\ \frac{1}{\sqrt{1 - (\frac{x}{\sqrt{5}})^2}}$
Now, you can substitute sin theta for x over root 5, which is what I've done.
Or you could sub root 5 sin theta for x.
for constants $\displaystyle a$ , $\displaystyle b$ , and $\displaystyle c$ ...
$\displaystyle \int \frac{a}{\sqrt{b - cx^2}} \, dx$
$\displaystyle \int \frac{a}{\sqrt{b} \cdot \sqrt{1 - \frac{c}{b}x^2}} \, dx$
$\displaystyle \frac{a}{\sqrt{b}} \int \frac{1}{\sqrt{1 - \frac{c}{b}x^2}} \, dx$
$\displaystyle \frac{a}{\sqrt{b}} \int \frac{1}{\sqrt{1 - \left(\sqrt{\frac{c}{b}}x \right)^2}} \, dx$
$\displaystyle \sqrt{\frac{b}{c}} \cdot \frac{a}{\sqrt{b}} \int \frac{\sqrt{\frac{c}{b}}}{\sqrt{1 - \left(\sqrt{\frac{c}{b}}x \right)^2}} \, dx$
$\displaystyle \frac{a}{\sqrt{c}} \int \frac{\sqrt{\frac{c}{b}}}{\sqrt{1 - \left(\sqrt{\frac{c}{b}}x \right)^2}} \, dx$
$\displaystyle \frac{a}{\sqrt{c}} \arcsin\left(\sqrt{\frac{c}{b}}x \right) + C$