# Tricky Trig Substitution

I'm working a problem for Cosmology, and I need to perform this integral: $$\int_0^a\frac{da}{\sqrt{\frac{8\pi G \epsilon_0 a}{3c^2}-ka^2}}.$$

From a similar problem, I know: $$\int_0^a\frac{da}{\sqrt{2Aa-a^2}}=\sin^{-1}\left(\frac{a-A}{A}\right)+\frac{\pi}{2}.$$

Can someone help me see the substitution I need to make to get this to work?

EDIT: Resolved, I worked through it backwards.

• Why not put $A=\frac{4\pi G_{\epsilon_0}}{3c^2}$? – Simply Beautiful Art Mar 3 '17 at 2:16
• @SimplyBeautifulArt: Rather, it should be $A=\frac{4\pi G\epsilon_0}{3c^2k}$, to factor out $k$ under the radical (and therefore $\sqrt{k}$ outside the radical). – zipirovich Mar 3 '17 at 2:20
• A small remark on the side, try to avoid having $a$ both in the bounds of integration and as a variable inside. – zwim Mar 3 '17 at 2:24

Well, first of all I set:

$$\text{n}=\frac{8\pi\text{G}\epsilon_0}{3\text{c}^2}\tag1$$

So, for the integral we get:

$$\mathscr{L}\space_\text{k}\left(\text{a}\right):=\int_0^\text{a}\frac{1}{\sqrt{\text{n}\text{a}-\text{k}\text{a}^2}}\space\text{d}\text{a}\tag2$$

Now, substitute $\text{u}=\text{a}\sqrt{\text{k}}-\frac{\text{n}}{2\sqrt{\text{k}}}$:

$$\mathscr{L}\space_\text{k}\left(\text{a}\right)=\frac{1}{\sqrt{\text{k}}}\int_{-\frac{\text{n}}{2\sqrt{\text{k}}}}^{\text{a}\sqrt{\text{k}}-\frac{\text{n}}{2\sqrt{\text{k}}}}\frac{1}{\sqrt{\frac{\text{n}^2}{4\text{k}}-\text{u}^2}}\space\text{d}\text{u}\tag3$$

Now, substitute $\text{s}=\frac{2\sqrt{\text{k}}\text{u}}{\text{n}}$:

$$\mathscr{L}\space_\text{k}\left(\text{a}\right)=\frac{1}{\sqrt{\text{k}}}\int_{-1}^{\frac{2\text{a}\text{k}}{\text{n}}-1}\frac{1}{\sqrt{1-\text{s}^2}}\space\text{d}\text{s}=\frac{1}{\sqrt{\text{k}}}\cdot\left[\arcsin\left(\text{s}\right)\right]_{-1}^{\frac{2\text{a}\text{k}}{\text{n}}-1}\tag3$$

So, we get:

$$\mathscr{L}\space_\text{k}\left(\text{a}\right)=\frac{\frac{\pi}{2}+\arcsin\left(\frac{2\text{a}\text{k}}{\text{n}}-1\right)}{\sqrt{\text{k}}}\tag4$$