Asymptotic approximation for $$\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\, dx$$ as $a \rightarrow 0$.
In such an integral, when $a$ approaches 0, the term $a^2x^2$ is always << 1 because x is bounded from $0$ to $1$. Thus, can I directly expand this term $\frac{1}{\sqrt{1-a^2x^2}} \sim (1+\frac{a^2x^2}{2}+O(a^4x^4))$ and integrate term by term multiplying the $\sqrt{1-x^2}$. Would doing this give me a good approximation for the integral? Thanks
Yes, with such approximation we find that as $a\to 0$, $$\int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\,dx= \frac{\pi}{4}+\frac{\pi a^2}{32}+O(a^4).$$ You may also obtain a more precise approximation, by using Wallis' integral, $$\int^1_0 \sqrt{1-x^2}\, x^{2n}\,dx=\int^{\pi/2}_0 \cos^2(t)\, \sin^{2n}(t)\,dt=\frac{\pi}{4^{n+1}(n+1)}\binom{2n}{n} .$$ Therefore, as $a\to 0$, $$\begin{align} \int^1_0 \frac{\sqrt{1-x^2}}{\sqrt{1-a^2x^2}}\,dx &=\int^1_0 \sqrt{1-x^2}\sum_{k=0}^{2n}\binom{-\frac{1}{2}}{k}(-a^2x^2)^{k}\,dx+O(a^{2n+2}) \\ &= \frac{\pi}{4}\sum_{k=0}^{2n}\frac{\binom{2n}{n}^2}{16^{n}(n+1)}\, a^{2k}+O(a^{2n+2}). \end{align}$$
