I need to choose a fitting change of variable to evaluate this integral: $$\phi=\phi_0 \mp \int^u du \left( \frac{2mE}{l^2} + \frac{2m^2\gamma u}{l^2}-u^2 \right)^{-1/2}$$ Is there any clever way of knowing what would be a good change of variable in a function such as this?
My initial instinct was to choose the entire term inside the parentheses as a substitution , but that doesn't seem correct at all. My guess to why that doesn't work is because the term includes both $u$ and $u^2$, but I can't explain why that is wrong. Can anybody help here as well?
Edit: In my book it says that "a standard integration gives $$\phi = \phi_0 \pm arccos \frac{1-ul^2/m^2\gamma}{(1+2El^2/m^3\gamma^2)^{1/2}}"$$
I assume the integral you are trying to solve is $$\int \left( \frac{2mE}{l^2} + \frac{2m^2\gamma u}{l^2}-u^2 \right)^{-1/2} \; \mathrm d u$$ because the upper bound $u$ does not make sence. Let us set $a=\frac{2mE}{l^2}$ and $b=\frac{2m^2\gamma}{l^2}$ where I assume $a\neq 0$ and $b\neq 0$. Therefore we are trying to solve the integral $$\int \frac{\mathrm d u}{\sqrt{a+bu-u^2}}.$$ The idea is to rewrite the integral as $$\text{const} \int \frac{1}{\sqrt{1-u^2 }} \mathrm d u = \text{const} \arcsin u.$$ We have $a+bu-u^2=a +\frac{1}{4}b^2-(u-\frac{1}{2}b)^2$ and therefore $$\int \frac{\mathrm d u}{\sqrt{a+bu-u^2}}= \int \frac{\mathrm d u}{\sqrt{a +\frac{1}{4}b^2-(u-\frac{1}{2}b)^2}} =\int \frac{1}{\sqrt{a +\frac{1}{4}b^2}}\frac{\mathrm d u}{\sqrt{1-\frac{(u-\frac{1}{2}b)^2}{a +\frac{1}{4}b^2}}} $$ $$=\frac{1}{\sqrt{a +\frac{1}{4}b^2}} \arcsin \left( 1-\frac{u-\frac{1}{2}b}{\sqrt{a +\frac{1}{4}b^2}} \right).$$ Resubstituting finally gives $$\int \left( \frac{2mE}{l^2} + \frac{2m^2\gamma u}{l^2}-u^2 \right)^{-1/2} \; \mathrm d u= \frac{1}{\sqrt{\frac{2mE}{l^2} +\left(\frac{2m^2\gamma}{2l^2}\right)^2}} \arcsin \left( 1-\frac{u-\frac{m^2\gamma}{l^2}}{\sqrt{\frac{2mE}{l^2} +\left(\frac{2m^2\gamma}{2l^2} \right)^2}} \right).$$