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Question Number 57138 by mustakim420 last updated on 30/Mar/19

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

Commented bymaxmathsup by imad last updated on 30/Mar/19

$$letf\left(a\right)={\int }_0^{\pi }\frac{dx}{a^2-2acosx+1}\Rightarrow f\left(a\right){=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^{+\mathrm{\infty }}\frac{2dt}{(1+t^2)(a^2-2a\frac{1-t^2}{1+t^2}+1)}$$ $$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{(1+t^2)a^2-2a(1-t^2)+1+t^2}=2{\int }_0^{\mathrm{\infty }}\frac{dt}{a^2{t}^2+a^2-2a+2{at}^2+1+t^2}$$ $$=2{\int }_0^{\mathrm{\infty }}\frac{dt}{(a^2+2a+1)t^2+a^2-2a+1}=2{\int }_0^{\mathrm{\infty }}\frac{dt}{{(a+1)}^2{t}^2+{(a-1)}^2}$$ $$changement(a+1)t=(a-1)ugive$$ $$f\left(a\right)=2{\int }_0^{\mathrm{\infty }}\frac{1}{{(a-1)}^2(1+u^2)}\frac{a-1}{a+1}du=\frac{2}{a^2-1}{\left[arctan\left(u\right)\right]}_0^{+\mathrm{\infty }}$$ $$=\frac{2}{a^2-1}\frac{\pi }{2}=\frac{\pi }{a^2-1}\Rightarrow f\left(a\right)=\frac{\pi }{a^2-1}witha>1.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Mar/19

$$\int \frac{dx}{a^2+1-2acosx}$$ $$\frac{1}{2a}\int \frac{dx}{k-cosx}[k=\frac{a^2+1}{2a}]$$ $$t=tan\frac{x}{2}2dt={sec}^2\frac{x}{2}dx$$ $$dx=\frac{2dt}{1+t^2}$$ $$\frac{1}{2a}\int \frac{2dt}{(1+t^2)(k-\frac{1-t^2}{1+t^2})}$$ $$\frac{1}{a}\int \frac{dt}{k+{kt}^2-1+t^2}$$ $$\frac{1}{a}\int \frac{dt}{(k-1)+t^2(k+1)}$$ $$\frac{1}{a(k+1)}\int \frac{dt}{\frac{k-1}{k+1}+t^2}$$ $$\frac{1}{a(\frac{a^2+1}{2a}+1)}\times \frac{1}{\sqrt{\frac{k-1}{k+1}}}{tan}^{-1}\left(\frac{t}{\sqrt{\frac{k-1}{k+1}}}\right)$$ $$$$ $$k=\frac{a^2+1}{2a}$$ $$\frac{k-1}{k+1}=\frac{a^2+1+2a}{a^2+1-2a}=\frac{{(a+1)}^2}{{(a-1)}^2}$$ $$\sqrt{\frac{k-1}{k+1}}=\frac{a+1}{a-1}$$ $$\frac{2}{{(a+1)}^2}\times \frac{1}{\frac{a-1}{a+1}}\mid {tan}^{-1}\left(\frac{tan\frac{x}{2}}{\frac{a-1}{a+1}}\right){\mid }_0^{\pi }$$ $$\frac{2}{a^2-1}\times [{tan}^{-1}\left(\mathrm{\infty }\right)-0]$$ $$\frac{2}{a^2-1}\times \frac{\pi }{2}\to \frac{\pi }{a^2-1}$$

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