Two integrals look almost the same, and even to those fairly wellversed in the art, are the same. But alas, as we shall see. Consider the integral $$I_1 = \int_0^{\pi} dx \frac{x \sin{x}}{1+\cos^2{x}} $$ This may be evaluated by subbing $x \mapsto \pix$ as follows: $$\begin{align} I_1 &= \int_0^{\pi} dx \frac{(\pix) \sin{x}}{1+\cos^2{x}} \\&= \pi \int_0^{\pi} […]

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