slick way of transforming an integral? 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?
 A: 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, $$I(a,b) = I(b,a)$$
A: I guess one way would be to go half-way though the Wierstrass substitution. Let $ \tan \frac{\zeta}{2} = t \tan \frac{\beta}{2}$, making it:
$$\begin{eqnarray}
   \int_0^\beta \frac{\sin \alpha}{1+\cos \alpha \cos \zeta} \mathrm{d}\zeta &=& \int_0^1 \frac{2 \tan \frac{\alpha}{2} }{\left(1+\tan^2 \frac{\alpha}{2}\right) + \left(1-\tan^2 \frac{\alpha}{2}\right) \frac{1- t^2 \tan^2 \frac{\beta}{2}}{1+ t^2 \tan^2 \frac{\beta}{2}} } \frac{2 \tan \frac{\beta}{2}}{1+ t^2 \tan^2 \frac{\beta}{2}} \mathrm{d}t \\
&=& \int_0^1 \frac{4 \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}{ \left(1 + \tan^2 \frac{\alpha}{2}\right)\left(1 + t^2 \tan^2 \frac{\beta}{2} \right) + \left(1 - \tan^2 \frac{\alpha}{2}\right)\left(1 - t^2 \tan^2 \frac{\beta}{2} \right)} \mathrm{d}t \\
 &=& \int_0^1 \frac{2 \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}{ 1 + t^2 \tan^2 \frac{\alpha}{2} \cdot \tan^2 \frac{\beta}{2} } \mathrm{d}t
\end{eqnarray}
$$
which is explicitly symmetric. However the closed form is now readily read off as a table integral:
$$
   \int_0^\beta \frac{\sin \alpha}{1+\cos \alpha \cos \zeta} \mathrm{d}\zeta = 2 \operatorname{arctan} \left( \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2} \right)
$$

Alternatively one could proceed with the differentiation. Let $F(\alpha,\beta)$ denote the original integral. Then
$$
   \frac{\partial F(\alpha,\beta)}{\partial \alpha} = \int_0^\beta \frac{\cos \alpha + \cos \zeta}{\left(1+\cos \alpha \cos \zeta\right)^2} \mathrm{d}\zeta = \left. \frac{\sin \zeta}{1+ \cos \alpha \cos \zeta} \right|_{\zeta=0}^{\zeta=\beta} = \frac{\sin \beta}{1 +\cos \alpha \cos \beta} = \frac{\partial F(\alpha,\beta)}{\partial \beta}
$$
This implies that $F(\alpha,\beta) - F\left(\beta,\alpha\right) = \mathrm{const.}$. The constant must be zero, since the constant is independent of $\alpha$ and $\beta$, and must equal to the negative of itself by interchangeing $\alpha$ and $\beta$ in the left-hand-side.
