TITLE: slick way of transforming an integral?
QUESTION [14 upvotes]: The function
$$
(\alpha,\beta) \mapsto \int_0^\beta \frac{\sin\alpha\,d\zeta}{1+\cos\alpha\cos\zeta}
$$
is a symmetric function of $\alpha$ and $\beta$.  But I don't know a simpler way to see that than by actually finding the integral.  Do the Weierstrass tangent half-angle substitution and keep turning the crank until you're there.  Suppose you don't want to know the integral but only want to show symmetry in $\alpha$ and $\beta$.  Is there some clever way to transform the integral into one in which $\alpha$ and $\beta$ play self-evidently symmetrical roles?

REPLY [9 votes]: First note that $$\dfrac{\sin(a)}{1+\cos(a) \cos(y)} = \left. \dfrac{\sin(x)}{1+ \cos(x) \cos(y)} \right \vert_{x=0}^{x=a}$$
Hence, let us try to write $\dfrac{\sin(a)}{1+\cos(a) \cos(y)}$ as an integral in $x$ from $0$ to $a$. We get that $$\dfrac{d}{dx} \left( \dfrac{\sin(x)}{1+ \cos(x) \cos(y)}\right) = \dfrac{\cos(x) + \cos(y)}{(1+\cos(x) \cos(y))^2}$$
Hence, we have that $$\dfrac{\sin(a)}{1+\cos(a) \cos(y)} = \int_{x=0}^{x=a}\dfrac{\cos(x) + \cos(y)}{(1+\cos(x) \cos(y))^2}dx$$
Hence, we have that
$$I(a,b) = \int_{y=0}^{y=b} \left(\int_{x=0}^{x=a}\dfrac{\cos(x) + \cos(y)}{(1+\cos(x) \cos(y))^2}dx \right) dy$$which is symmetric about $a$ and $b$. Hence, $$I(a,b) = I(b,a)$$