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Question Number 36197 by prof Abdo imad last updated on 30/May/18

$$findthevalueof{\int }_0^{2\pi }\frac{dx}{{cos}^{2}x+3{sin}^{2}x}$$

Commented by maxmathsup by imad last updated on 14/Aug/18

$$letA={\int }_0^{2\pi }\frac{dx}{{cos}^{2}x+3{sin}^{2}x}$$$$A={\int }_0^{2\pi }\frac{dx}{\frac{1+cos\left(2x\right)}{2}+\frac{1-sin\left(2x\right)}{2}}={\int }_0^{2\pi }\frac{2dx}{2+cos\left(2x\right)-sin\left(2x\right)}$$$${=}_{2x=t}{\int }_0^{4\pi }\frac{dt}{2+cost-sint}={\int }_0^{2\pi }\frac{dt}{2+cost-sint}+{\int }_{2\pi }^{4\pi }\frac{dt}{2+cost-sint}$$$$but{\int }_{2\pi }^{4\pi }\frac{dt}{2+cost-sint}{=}_{t=2\pi +\alpha }{\int }_0^{2\pi }\frac{d\alpha }{2+cos\alpha -sin\alpha }\Rightarrow$$$$A={\int }_0^{2\pi }\frac{2dt}{2+cost-sint}changement{e}^{it}=zgive$$$$A={\int }_{\mid z\mid =1}\frac{2}{2+\frac{z+z^{-1}}{2}+\frac{z-z^{-1}}{2i}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{4dz}{iz(4+z+z^{-1}-i(z-z^{-}))}={\int }_{\mid z\mid =1}\frac{-4idz}{4z+z^2+1-{iz}^2+i}$$$$={\int }_{\mid z\mid =1}\frac{-4iz}{(1-i)z^2+4z+1+i}letconsiderthecompolexfunction$$$$\phi \left(z\right)=\frac{-4i}{(1-i)z^2+4z+1+i}polesof\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}=4-(1-i)(1+i)=4-2=2\Rightarrow z_1=\frac{-2+\sqrt{2}}{1-i}=\frac{\sqrt{2}(1-\sqrt{2})}{\sqrt{2}{e}^{-\frac{i\pi }{4}}}=(1-\sqrt{2})e^{\frac{i\pi }{4}}$$$$z_2=\frac{-2-\sqrt{2}}{1-i}=\frac{\sqrt{2}(-1-\sqrt{2})}{\sqrt{2}{e}^{-\frac{i\pi }{4}}}=(-1-\sqrt{2})e^{\frac{i\pi }{4}}$$$$\mid z_1\mid =\sqrt{2}-1<1and\mid z_2\mid =\sqrt{2}+1>1\left(toeliminatefromresidus\right)\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)but\phi \left(z\right)=\frac{-4i}{(1-i)(z-z_1)(z-z_2)}$$$$Res(\phi ,z_1)=\frac{-4i}{(1-i)(z_1-z_2)}=\frac{-4i}{(1-i)2e^{i\frac{\pi }{4}}}=\frac{-2i}{\sqrt{2}}=\frac{-i}{\sqrt{2}}\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi \left(\frac{-i}{\sqrt{2}}\right)=\frac{2\pi }{\sqrt{2}}\Rightarrow A=\frac{2\pi }{\sqrt{2}}\Rightarrow A=\pi \sqrt{2}.$$

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