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Question Number 27295 by iy last updated on 04/Jan/18

$$\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{integral}$$$$\underset{0}{\stackrel{\pi }{\int }}\frac{1}{a^2-2a\cos x+1}dx(a<1)\mathrm{is}$$

Commented by abdo imad last updated on 04/Jan/18

$$letdothechangementtan\left(\frac{x}{2}\right)=t$$$$I={\int }_0^{\pi }\frac{dx}{a^2-2acosx+1}={\int }_0^{\propto }\frac{\frac{dt}{1+t^2}}{a^2-2a\frac{1-t^2}{1+t^2}+1}$$$$$$$$={\int }_0^{\propto }\frac{dt}{a^2(1+t^2)-2a(1-t^2)+1+t^2}$$$$={\int }_0^{\propto }\frac{dt}{(a^2+2a+1)t^2+a^2-2a+1}$$$$={\int }_0^{\propto }\frac{dt}{{(a+1)}^2{t}^2+{(a-1)}^2}$$$$={\int }_0^{\propto }\frac{dt}{{(a+1)}^2(t^2+{\left(\frac{1-a}{a+1}\right)}^2)}andwedothechangeent$$$$t=\frac{1-a}{a+1}\alpha \Rightarrow I=\frac{1}{{(a+1)}^2}{\int }_0^{\propto }\frac{\frac{1-a}{a+1}d\alpha }{{\left(\frac{1-a}{a+1}\right)}^2(1+{\alpha }^2)}$$$$=\frac{1}{{(a+1)}^2}\frac{1-a}{a+1}\frac{{(a+1)}^2}{{(1-a)}^2}\frac{\pi }{2}$$$$=\frac{\pi }{2(1-a^2)}.$$

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