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Question Number 200309 by sonukgindia last updated on 17/Nov/23

Answered by Mathspace last updated on 17/Nov/23

$$J={\int }_0^{2\pi }\frac{dx}{a^2-2asinx+1}$$$$={\int }_0^{2\pi }\frac{dx}{a^2-2a\frac{e^{ix}-e^{-ix}}{2i}+1}(e^{ix}=z)$$$$={\int }_{\mid z\mid =1}\frac{dz}{iz(a^2+ia(z-z^{-1})+1)}$$$$={\int }_{\mid z\mid =1}\frac{-idz}{a^{2}z+{iaz}^2-ia+z}$$$$={\int }_{\mid z\mid =1}\frac{-idz}{{iaz}^2+(a^2+1)z-ia}$$$$f\left(z\right)=\frac{-i}{{iaz}^2+(a^2+1)z-ia}$$$$polesoff?$$$$\mathrm{\Delta }={(a^2+1)}^2-4\left(ia\right)(-ia)$$$$=a^4+2a^2+1-4a^2={(a^2-1)}^2$$$$and\sqrt{\mathrm{\Delta }}=\mid a^2-1\mid =1-a^2$$$$z_1=\frac{-a^2-1+1-a^2}{2ia}=-\frac{1}{i}a$$$$=ia\Rightarrow \mid z_1\mid =\mid a\mid <1$$$$z_2=\frac{-a^2-1-1+a^2}{2ia}=-\frac{1}{ia}$$$$=\frac{i}{a}\Rightarrow \mid z_2\mid =\frac{1}{a}>1$$$$residustheoremgive$$$${\int }_{\mid z\mid =1}f\left(z\right)dz=2i\pi Res(f,z_1)$$$$Res(f,z_1)={lim}_{z\to z_1}(z-z_1)\frac{-ia}{ia(z-z_1)(z-z_2)}$$$$=\frac{-1}{z_1-z_2}=\frac{-1}{ia-\frac{i}{a}}=\frac{i}{a^2-1}$$$$\Rightarrow {\int }_{\mid z\mid =1}f\left(z\right)dz=2i\pi .\frac{i}{a^2-1}$$$$=\frac{-2\pi }{a^2-1}=\frac{2\pi }{1-a^2}\Rightarrow$$$$\bullet J=\frac{2\pi }{1-a^2}\bullet$$$$$$

Answered by Frix last updated on 17/Nov/23

$$\underset{0}{\stackrel{2\pi }{\int }}\frac{dx}{a^2-2a\sin x+1}=$$$$=2\underset{0}{\stackrel{\pi }{\int }}\frac{dx}{a^2-2a\cos x+1}\stackrel{t=\frac{1+a}{1-a}\tan \frac{x}{2}}{=}$$$$=\frac{4}{1-a^2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t^2+1}=\frac{2\pi }{1-a^2}$$

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