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Question Number 130529 by mathmax by abdo last updated on 26/Jan/21

$$\mathrm{calvulste}{\int }_0^{2\pi }\frac{\cos \left(2\mathrm{x}\right)}{3+\mathrm{cosx}}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 27/Jan/21

$$\mathrm{A}={\int }_0^{2\pi }\frac{\cos \left(2\mathrm{x}\right)}{3+\mathrm{cosx}}\mathrm{dx}\Rightarrow \mathrm{A}={\int }_0^{2\pi }\frac{{2\cos }^2\mathrm{x}-1}{\mathrm{cosx}+3}\mathrm{dx}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{u}\right)=\frac{{2\mathrm{u}}^2-1}{\mathrm{u}+3}\Rightarrow \mathrm{F}\left(\mathrm{u}\right)=\frac{2({\mathrm{u}}^2-9)+18-1}{\mathrm{u}+3}$$$$=2(\mathrm{u}-3)+\frac{17}{\mathrm{u}+3}\Rightarrow \mathrm{A}={\int }_0^{2\pi }(2\mathrm{cosx}-6)\mathrm{dx}+17{\int }_0^{2\pi }\frac{\mathrm{dx}}{\mathrm{cosx}+3}$$$$=-12\pi +{\left[2\mathrm{sinx}\right]}_0^{2\pi }+17\mathrm{\Phi }=17\mathrm{\Phi }-12\pi$$$$\mathrm{\Phi }={\int }_0^{2\pi }\frac{\mathrm{dx}}{\mathrm{cosx}+3}{=}_{{\mathrm{e}}^{\mathrm{ix}}=\mathrm{z}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2}+3)}$$$$={\int }_{\mid \mathrm{z}\mid =1}\frac{2\mathrm{dz}}{\mathrm{iz}(\mathrm{z}+{\mathrm{z}}^{-1}+6)}=-2\mathrm{i}{\int }_{\mid \mathrm{z}\mid }\frac{\mathrm{dz}}{{\mathrm{z}}^2+1+6\mathrm{z}}=\int \frac{-2\mathrm{idz}}{{\mathrm{z}}^2+6\mathrm{z}+1}$$$$\phi \left(\mathrm{z}\right)=\frac{-2\mathrm{i}}{{\mathrm{z}}^2+6\mathrm{z}+1}\mathrm{poles}\mathrm{of}\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}=9-1=8\Rightarrow {\mathrm{z}}_1=-3+2\sqrt{2}\mathrm{and}{\mathrm{z}}_2=-3-2\sqrt{2}\mathrm{and}\phi \left(2\right)=\frac{-2\mathrm{i}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}$$$$\mid {\mathrm{z}}_1\mid -1=\mid -3+2\sqrt{2}\mid -1=3-2\sqrt{2}-1=2-2\sqrt{2}<0\Rightarrow \mid {\mathrm{z}}_1\mid <1$$$$\mid {\mathrm{z}}_2\mid >1\left(\mathrm{out}\mathrm{of}\mathrm{circle}\right)\Rightarrow {\int }_{\mid \mathrm{z}\mid =1}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,{\mathrm{z}}_1)$$$$=2\mathrm{i}\pi \times \frac{-2\mathrm{i}}{{\mathrm{z}}_1-{\mathrm{z}}_2}=\frac{4\pi }{4\sqrt{2}}=\frac{\pi }{\sqrt{2}}\Rightarrow \mathrm{\Phi }=\frac{\pi }{\sqrt{2}}\Rightarrow$$$$\mathrm{A}=\frac{17\pi }{\sqrt{2}}-12\pi =(\frac{17}{\sqrt{2}}-12)\pi$$$$$$

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