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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?

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  • $\begingroup$ It has been pointed out in this forum by Fred Rickey that Weiestrass had nothing to do with that subsitution that Stewart's calculus text names after him, so I've replaced that name. $\endgroup$ – Michael Hardy Dec 27 '13 at 17:48
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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, $$I(a,b) = I(b,a)$$

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  • $\begingroup$ I'm already wondering if this is a special case of something involving a fraction whose numerator is $e_1+e_3+e_5+\cdots$ where $e_k$ is the $k$th-degree elementary symmetric function in $\cos(x_1),\cos(x_2),\cos(x_3),\ldots$, and whose denominator is the square of $e_0+e_2+e_4+\cdots$.${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 27 '12 at 15:04
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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.

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  • $\begingroup$ or you could set $\beta = 0$ to get that $F(\alpha, 0) = F(0,\alpha) = 0 -0 = 0$. $\endgroup$ – user17762 Dec 27 '12 at 4:43

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