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\int_0^{\pi}\frac{ad\theta}{a^2 + \sin^2\theta} = \int_0^{2\pi}\frac{ad\theta}{1 + 2a^2 - \cos\theta} = \frac{\pi}{\sqrt{1 + a^2}}

$$

Consider $a > 0$ and $a < 0$

First I don't think the second part is correct. Shouldn't it be $1 + 2a^2 - \cos 2\theta$?