Integral of a trig function divided by the square root of a polynomial: $\int_a^b\frac{\sin x}{\sqrt{(x-a)(b-x)}}dx$?

Question

I was trying to help some physics students with an integral on their homework and they've presented me with something that has me stumped.

The integral they are working on is:

$$\int_a^b\frac{\sin x}{\sqrt{(x-a)(b-x)}}dx$$

They're definitely going to need a change of variable but any changes to simplify the bottom portion will greatly complicate the top. Any suggestions on how to tackle such a beast?

Answer 1

Rewrite as

$$\begin{align}\int_0^{b-a} dx \frac{\sin{(x+a)}}{\sqrt{x (b-a-x)}} &= \int_0^1 du \frac{\sin{[(b-a) u+a]}}{\sqrt{u (1-u)}}\\ &= \cos{a}\int_0^1 du \frac{\sin{[(b-a) u]}}{\sqrt{u (1-u)}} + \sin{a} \int_0^1 du \frac{\cos{[(b-a) u]}}{\sqrt{u (1-u)}}\\ &= 2 \cos{a} \int_0^1 du \frac{\sin{[ (b-a) u^2 ]}}{\sqrt{1-u^2}} + 2 \sin{a} \int_0^1 du \frac{\cos{[ (b-a) u^2 ]}}{\sqrt{1-u^2}}\end{align}$$

Now, consider

$$\begin{align}\int_0^1 du \, (1-u^2)^{-1/2} e^{i (b-a) u^2} &= \int_0^{\pi/2} dt \, e^{i (b-a) \sin^2{t}}\\ &= e^{i (b-a)/2} \int_0^{\pi/2} dt \, e^{-i [(b-a)/2] \cos{2 t}}\\ &= \frac12 e^{i (b-a)/2} \int_0^{\pi} dt \, e^{-i [(b-a)/2] \cos{t}}\\ &= \frac{\pi}{2} e^{i (b-a)/2} J_0 \left ( \frac{b-a}{2}\right ) \end{align}$$

Putting this altogether, the integral is

$$\pi J_0 \left ( \frac{b-a}{2}\right ) \left [ \cos{a} \sin{ \left ( \frac{b-a}{2}\right )} + \sin{a} \cos{ \left ( \frac{b-a}{2}\right )}\right ] $$

or

$$\int_a^b dx \frac{\sin{x}}{\sqrt{(x-a)(b-x)}} = \pi J_0 \left ( \frac{b-a}{2}\right ) \sin{ \left ( \frac{a+b}{2}\right )}$$