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Question Number 28815 by abdo imad last updated on 30/Jan/18

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

Answered by mrW2 last updated on 31/Jan/18

$${\int }_0^{\pi }\frac{dx}{2{cos}^{2}x+{sin}^{2}x}$$$$={\int }_0^{\pi }\frac{dx}{1+{cos}^{2}x}$$$$=2{\int }_0^{\pi }\frac{dx}{3+2{cos}^{2}x-1}$$$$=2{\int }_0^{\pi }\frac{dx}{3+\cos 2x}$$$$=2\times 2{\left[\frac{{\tan }^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)}{2\sqrt{2}}\right]}_0^{\pi /2}$$$$=\frac{\pi }{\sqrt{2}}$$

Commented by abdo imad last updated on 31/Jan/18

$$anothermetodbyredidustheoremwehave$$$$I={\int }_0^{\pi }\frac{dx}{1+{cos}^{2}x}={\int }_0^{\pi }\frac{dx}{1+\frac{1+cos\left(2x\right)}{2}}={\int }_0^{\pi }\frac{2dx}{3+cos\left(2x\right)}$$$$thech.2x=tgiveI={\int }_0^{2\pi }\frac{dt}{3+cost}andthech.e^{it}=zgive$$$$I={\int }_{\mid z\mid =1}\frac{1}{3+\frac{z+z^{-1}}{2}}\frac{dz}{iz}={\int }_{\mid z\mid =1}\frac{2dz}{iz(6+z+z^{-1})}$$$$={\int }_{\mid z\mid }\frac{-2i}{6z+z^2+1}={\int }_{\mid z\mid =1}\frac{-2i}{z^2+6z+1}dzletput$$$$f\left(z\right)=\frac{-2i}{z^2+6z+1},polesoff?$$$$z^2+6z+1=0\Rightarrow \mathrm{\Delta }=36-4=32\Rightarrow z_1=\frac{-6+4\sqrt{2}}{2}=-3+2\sqrt{2}$$$$z_2=-3-2\sqrt{2}wehave\mid z_1\mid -1=2\sqrt{2}-3-1=2(\sqrt{2}-2)<0$$$$and\mid z_2\mid -1=3+2\sqrt{2}-1=2+2\sqrt{2}>1(toeliminatebecause$$$$outofcircle)$$$${\int }_{\mid z\mid =1}f\left(z\right)dx=2i\pi Re(f,z_1)butf\left(z\right)=\frac{-2i}{(z-z_1)(z-z_2)}$$$$Res(f,z_1)={lim}_{z\to z_1}(z-z_1)f\left(z\right)=\frac{-2i}{z_1-z_2}=\frac{-2i}{4\sqrt{2}}$$$${\int }_{\mid z\mid =1}f\left(z\right)dx=2i\pi .\frac{-i}{2\sqrt{2}}=\frac{\pi }{\sqrt{2}}soI=\frac{\pi }{\sqrt{2}}.$$

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