When $a\ll b$, how to approximate $f = \int_0^a \sqrt{b^2+x^2}/\sqrt{a^2-x^2} \, \, dx$?

Question

Suppose $a\ll b$. How do I then approximate $$\int_0^a \frac{\sqrt{b^2+x^2}}{\sqrt{a^2-x^2}}dx$$ ?

I think that maybe Taylor approximation may help, but I am not sure how to proceed.

My physics textbook says that the answer is $\pi b [1+a^2/(4b^2)]/2 $.

Answer 1

Let $x=a\sin t \quad t\in[0,\frac{\pi}{ 2}]$ ,and $\sqrt{1+x^2}\sim 1+\frac{x^2}{2}$, when $x$ is small.

Then $\int_0^a \frac{\sqrt{b^2+x^2}}{\sqrt{a^2-x^2}}dx=\int_0^{\frac{\pi}{ 2}} \sqrt{b^2+a^2\sin^2(t)}dt=\int_0^{\frac{\pi}{ 2}} b\sqrt{1+\frac{a^2}{b^2}\sin^2(t)}dt\sim \int_0^{\frac{\pi}{ 2}} b(1+\frac{a^2}{2b^2}\sin^2(t))dt$

Note the assumption $a\ll b$ makes this approximation valid. Then use double angle formula and evaluate the integral, you will get the result.

Answer 2

Depending on the problem, there are two solutions to approach the approximation.

One has been already addressed by John ZHANG in his answer : approximate the integrand and integrate.

Another one is to compute, if feasible, the integral and approximate the result. In the case you posted, we have $$I=\int_0^a \frac{\sqrt{b^2+x^2}}{\sqrt{a^2-x^2}}dx=b E\left(-\frac{a^2}{b^2}\right)$$ where appears the elliptic integral of the second kind. Expanding the result as a Taylor series, we then have $$I=\frac{\pi b}{2}\Big(1+\frac{ a^2}{4 b^2}-\frac{3 a^4}{64 b^4}+\frac{5 a^6}{256 b^6}+O\left(a^7\right)\Big)$$