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Question Number 188511 by universe last updated on 02/Mar/23

$$evaluate$$$${\int }_0^{\pi }\frac{dx}{a+b\cos x},a>0$$$$anddeducethat$$$${\int }_0^{\pi }\frac{dx}{{(a+b\cos x)}^2}=\frac{\pi a}{{(a^2-b^2)}^{3/2}};a^2>b^2$$$$and{\int }_0^{\pi }\frac{\cos xdx}{{(a+b\cos x)}^2}=\frac{-\pi b}{{(a^2-b^2)}^{3/2}};a^2>b^2$$

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