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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 $$\phi = \phi_0 \pm arccos \frac{1-ul^2/m^2\gamma}{(1+2El^2/m^3\gamma^2)^{1/2}}"$$

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    Express the parabola in vertex form, then it might become more clear. Do you know any trig substitutions? – 2017-02-12
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    The upper bound is $u$? – 2017-02-12
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    @NiklasHebestreit, yes that is how it's written. We're talking about an integral related to Kepler's laws – 2017-02-12
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    Well, then my answer is not usefull i guess. How can the upper bound depend on the integration variable? – 2017-02-12
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    @NiklasHebestreit, so far I've ignored the upper bound as well. The integral is itself a substitution from another integral, but all the way through the upper bound is the variable being integrated. Your answer looks like what I'm trying to get to - I've just edited my question so that it includes what I'm supposed to end up with. – 2017-02-12

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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 $$\int \left( \frac{2mE}{l^2} + \frac{2m^2\gamma u}{l^2}-u^2 \right)^{-1/2} \; \mathrm d u= \frac{1}{\sqrt{\frac{2mE}{l^2} +\left(\frac{2m^2\gamma}{2l^2}\right)^2}} \arcsin \left( 1-\frac{u-\frac{m^2\gamma}{l^2}}{\sqrt{\frac{2mE}{l^2} +\left(\frac{2m^2\gamma}{2l^2} \right)^2}} \right).$$