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Question Number 37587 by prof Abdo imad last updated on 15/Jun/18

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

Commented by math khazana by abdo last updated on 16/Jun/18

$$I={\int }_0^{2\pi }\frac{dx}{2\frac{1+cos\left(2x\right)}{2}+\sqrt{3}\frac{1-cos\left(2x\right)}{2}}$$$$={\int }_0^{2\pi }\frac{dx}{1+cos\left(2x\right)+\frac{\sqrt{3}}{2}(1-cos\left(2x\right))}$$$$={\int }_0^{2\pi }\frac{2dx}{2+2cos\left(2x\right)+\sqrt{3}-\sqrt{3}cos\left(2x\right)}$$$$={\int }_0^{2\pi }\frac{2dx}{2+\sqrt{3}+(2-\sqrt{3})cos\left(2x\right)}$$$${=}_{2x=t}{\int }_0^{4\pi }\frac{2dx}{2+\sqrt{3}+(2-\sqrt{3})cost}\frac{dt}{2}$$$$={\int }_0^{4\pi }\frac{dt}{2+\sqrt{3}+(2-\sqrt{3})cost}$$$$={\int }_0^{2\pi }\frac{dt}{2+\sqrt{3}+(2-\sqrt{3})cost}+{\int }_{2\pi }^{4\pi }\frac{dt}{2+\sqrt{3}+(2-\sqrt{3})cost}$$$$=K+H$$$$chang.e^{it}=zgive$$$$K={\int }_{\mid z\mid =1}\frac{1}{2+\sqrt{3}+(2-\sqrt{3})\frac{z+z^{-1}}{2}}\frac{dz}{iz}$$$$={\int }_{\mid z\mid =1}\frac{2dz}{iz(4+2\sqrt{3}+(2-\sqrt{3})(z+z^{-1})}$$$$={\int }_{\mid z\mid =1}\frac{-2idz}{(2-\sqrt{3})z^2+(4+2\sqrt{3})z+2-\sqrt{3}}$$$$let\phi \left(z\right)=\frac{-2i}{(2-\sqrt{3})z^2+(4+2\sqrt{3})z+2-\sqrt{3}}$$$$polesof\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}={(2+\sqrt{3})}^2-{(2-\sqrt{3})}^2=4+4\sqrt{3}+3-4+4\sqrt{3}-3$$$$=8\sqrt{3}$$$$z_1=\frac{-2-\sqrt{3}+2\sqrt{2}\sqrt{\sqrt{3}}}{2-\sqrt{3}}$$$$z_2=\frac{-2-\sqrt{3}-2\sqrt{2}\sqrt{\sqrt{3}}}{2-\sqrt{3}}$$$$\mid z_1\mid -1=\frac{\mid 2+\sqrt{3}-2\sqrt{2}\sqrt{\sqrt{3}}\mid }{2-\sqrt{3}}$$$${(2+\sqrt{3})}^2-8\sqrt{3}=7-4\sqrt{3}-8\sqrt{3}=7-12\sqrt{3}<0$$$$\Rightarrow \mid z_1\mid <1andweverifythat\mid z_2\mid >1$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)$$$$Res(\phi ,z_1)=\frac{-2i}{(2-\sqrt{3})(z_1-z_2)}$$$$=\frac{-2i}{(2-\sqrt{3})\frac{4\sqrt{2\sqrt{3}}}{2-\sqrt{3}}}=\frac{-i}{2\sqrt{2\sqrt{3}}}.\Rightarrow$$$${\int }_{\mid z\mid =1}\phi \left(z\right)dz=2i\pi \frac{-i}{2\sqrt{2\sqrt{3}}}=\frac{\pi }{\sqrt{2\sqrt{3}}}=K.$$

Commented by math khazana by abdo last updated on 16/Jun/18

$$H={\int }_{2\pi }^{4\pi }\frac{dt}{2+\sqrt{3}+(2-\sqrt{3})cost}$$$${=}_{t=2\pi +x}{\int }_0^{2\pi }\frac{dx}{2+\sqrt{3}+(2-\sqrt{3})cosx}=K$$$$=\frac{\pi }{\sqrt{2\sqrt{3}}}\Rightarrow$$$$I=H+K=\frac{2\pi }{\sqrt{2\sqrt{3}}}=\frac{\pi \sqrt{2}}{\sqrt{\sqrt{3}}}$$$$I=\frac{\pi \sqrt{2}}{\sqrt{\sqrt{3}}}.$$

Answered by ajfour last updated on 15/Jun/18

$${\int }_0^{2\pi }\frac{\sec {}^{2}xdx}{2+\sqrt{3}\tan {}^{2}x}$$$$=\frac{4}{\sqrt{3}}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+\frac{2}{\sqrt{3}}}$$$$=\frac{4}{\sqrt{3}}\times \frac{3^{1/4}}{\sqrt{2}}\left(\frac{\pi }{2}\right)=\pi \sqrt{\frac{2}{\sqrt{3}}}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

$$inplaceof\frac{\mathrm{\Pi }}{2}iput\frac{\mathrm{\Pi }}{4}$$

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