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Question Number 36936 by maxmathsup by imad last updated on 07/Jun/18

$$calculate{I}_n={\int }_0^{\pi }\frac{dx}{1+{cos}^2\left(nx\right)}$$

Commented by math khazana by abdo last updated on 02/Aug/18

$$wehave{I}_n={\int }_0^{\pi }\frac{dx}{1+\frac{1+cos\left(2nx\right)}{2}}$$$$={\int }_0^{\pi }\frac{2dx}{3+cos\left(2nx\right)}{=}_{2nx=t}{\int }_0^{2n\pi }\frac{2}{3+cost}\frac{dt}{2n}$$$$=\frac{1}{n}{\int }_0^{2n\pi }\frac{dt}{3+cost}\Rightarrow {nI}_n=\sum _{k=0}^{n-1}{\int }_{2k\pi }^{2(k+1)\pi }\frac{dt}{3+cost}$$$${=}_{t=2k\pi +x}\sum _{k=0}^{n-1}{\int }_0^{2\pi }\frac{dx}{3+cos(x+2k\pi )}$$$$=n{\int }_0^{2\pi }\frac{dx}{3+cosx}\Rightarrow I_n={\int }_0^{2\pi }\frac{dx}{3+cosx}\mathrm{\forall }n⩾1and$$$$I_o=\pi changement{e}^{ix}=zgive$$$$I_n={\int }_{\mid z\mid =1}\frac{1}{3+\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz\{6+z+z^{-1}\}}={\int }_{\mid z\mid =1}\frac{-2idz}{6z+z^2+1}$$$$let\phi \left(z\right)=\frac{-2i}{z^2+6z+1}polesof\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}=3^2-1=8\Rightarrow z_1=-3+2\sqrt{2}and{z}_2=-3-2\sqrt{2}$$$$\mid z_1\mid -1=3-2\sqrt{2}-1=2-2\sqrt{2}<0\Rightarrow \mid z_1\mid <1$$$$\mid z_2\mid -1=3+2\sqrt{2}-1=2+2\sqrt{2}>1\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)but$$$$\phi \left(z\right)=\frac{-2i}{(z-z_1)(z-z_2)}\Rightarrow Res(\phi ,z_1)=\frac{-2i}{z_1-z_2}$$$$=\frac{-2i}{4\sqrt{2}}=\frac{-i}{2\sqrt{2}}\Rightarrow {\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi .\frac{-i}{2\sqrt{2}}=\frac{\pi }{\sqrt{2}}\Rightarrow$$$$I_n=\frac{\pi }{\sqrt{2}}withn⩾1.$$$$$$$$$$$$$$

Commented by math khazana by abdo last updated on 02/Aug/18

$$I_0=\frac{\pi }{2}$$

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