Help with integral $\int^{a\sqrt{1-E/V}}_{-a\sqrt{1-E/V}}\sqrt{2mV(1-x^2/a^2)-E} dx $ Hi I have tried u and trig substitution for this integral and just cant get it can someone offer a pointer or two? Thanks   
$\int^{a\sqrt{1-E/V_0}}_{-a\sqrt{1-E/V_0}}\sqrt{2m[V_0(1-x^2/a^2)-E]} dx $

One can simplify the integral using substitutions.
$$I=\int^{a\sqrt{1-E/V_0}}_{-a\sqrt{1-E/V_0}}\sqrt{2m(V_0(1-x^2/a^2)-E)} \ \mathrm dx=\sqrt{2mV_0}\int^{a\sqrt{1-E/V_0}}_{-a\sqrt{1-E/V_0}}\sqrt{1-\frac{E}{V_0}-\frac{x^2}{a^2}} \ \mathrm dx$$
Let's set $\sqrt{1-E/V_0}=\alpha$, then changing the variable to $t=\frac{x}{a}$ one gets 
$$I=\sqrt{2mV_0} a\int^{\alpha}_{-\alpha}\sqrt{\alpha^2-t^2} \ \mathrm dt=2a\sqrt{2 m V_0} \int^{\alpha}_{0}\sqrt{\alpha^2-t^2} \ \mathrm dt$$ since it is a symmetric integral of an even function.
And in cases when $\alpha>0$ $$\int^{\alpha}_{0}\sqrt{\alpha^2-t^2} \ \mathrm dt=\frac{\pi\alpha^2}{4}.$$
