Q: Differentiating under the integral If
$$
I(\alpha) = \int_{0}^{\pi}\frac{dx}{\alpha -\cos x} = \frac{\pi}{\sqrt{\alpha ^2-1}}
$$
where $$\alpha > 1$$
Then what is
$$\int_{0}^{\pi}\frac{dx}{(2-\cos x)^2}$$
?
 A: Given $$\displaystyle I(\alpha) = \int_{0}^{\pi}\frac{1}{\alpha-\cos x} = \frac{\pi}{\sqrt{\alpha^2-1}}\;,$$ Where $\alpha>1$
Now Differentiate both side w. r to $\alpha\;,$ We get
$$\displaystyle \frac{d}{d\alpha}\left[I(\alpha)\right] = \int_{0}^{\pi}\frac{d}{d\alpha}\left[\frac{1}{\alpha-\cos x}\right]dx = \frac{d}{d\alpha}\left[\frac{\pi}{\sqrt{\alpha^2-1}}\right]$$
Now $$\displaystyle I'(\alpha) = -\int_{0}^{\pi}\frac{1}{(\alpha-\cos x)^2}dx = -\frac{\pi}{2}\cdot \frac{1}{(\alpha^2-1)^{\frac{3}{2}}}\cdot 2\alpha$$
So we get $$\displaystyle \int_{0}^{\pi}\frac{1}{(\alpha-\cos x)^2}dx = \frac{\pi\cdot \alpha}{(\alpha^2-1)\sqrt{\alpha^2-1}}.$$
Now Put $\alpha=2\;,$ we get
$$\displaystyle \int_{0}^{\pi}\frac{1}{(2-\cos x)^2}dx = \frac{\pi\cdot 2}{(2^2-1)\sqrt{2^2-1}}=\frac{2\pi}{3\sqrt{3}}.$$
A: $$\frac{d}{d\alpha}\frac{1}{\alpha-\cos x}=-\frac{1}{(\alpha-\cos x)^2}$$
so
$$\int_{0}^{\pi}\frac{dx}{(\alpha-\cos x)^2}=-\pi \frac{d}{d\alpha} (\alpha^2-1)^{-\frac 12} = \pi \alpha (\alpha^2-1)^{-\frac 32} $$
