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Question Number 146371 by lapache last updated on 13/Jul/21

$${\int }_0^{\pi }\frac{1}{1+{cos}^{2}x}dx=......???$$

Answered by gsk2684 last updated on 13/Jul/21

$$letI=\underset{0}{\stackrel{\mathrm{\Pi }}{\int }}\frac{1}{1+\cos {}^{2}x}dx\left(1\right)$$$$f(\mathrm{\Pi }-x)=\frac{1}{1+\cos {}^2(\mathrm{\Pi }-x)}=\frac{1}{1+\cos {}^{2}x}=f\left(x\right)$$$$\left(1\right)\Rightarrow I=2\underset{0}{\stackrel{\frac{\mathrm{\Pi }}{2}}{\int }}\frac{1}{1+\cos {}^{2}x}dx$$$$\Rightarrow I=2\underset{0}{\stackrel{\frac{\mathrm{\Pi }}{2}}{\int }}\frac{1}{{\sec }^{2}x+1}\sec {}^{2}xdx$$$$\Rightarrow I=2\underset{0}{\stackrel{\frac{\mathrm{\Pi }}{2}}{\int }}\frac{1}{1+\tan {}^{2}x+1}\sec {}^{2}xdx$$$$\Rightarrow I=2\underset{0}{\stackrel{\frac{\mathrm{\Pi }}{2}}{\int }}\frac{1}{\tan {}^{2}x+{\left(\sqrt{2}\right)}^2}d\left(tanx\right)$$$$\Rightarrow I=2{\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)\right]}_0^{\frac{\mathrm{\Pi }}{2}}$$$$\Rightarrow I=\sqrt{2}[\frac{\mathrm{\Pi }}{2}-0]=\frac{\mathrm{\Pi }}{\sqrt{2}}$$

Answered by Ar Brandon last updated on 13/Jul/21

$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{dx}}{1+{\cos }^2\mathrm{x}}=2{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\mathrm{x}}{{\sec }^2\mathrm{x}+1}\mathrm{dx}$$$$=2{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{d}\left(\mathrm{tanx}\right)}{2+{\tan }^2\mathrm{x}}=\frac{2}{\sqrt{2}}{\left[\arctan \left(\frac{\mathrm{tanx}}{\sqrt{2}}\right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{\sqrt{2}}$$

Answered by mathmax by abdo last updated on 13/Jul/21

$$\mathrm{\Psi }={\int }_0^{\pi }\frac{\mathrm{dx}}{1+{\cos }^2\mathrm{x}}\Rightarrow \mathrm{\Psi }={\int }_0^{\pi }\frac{\mathrm{dx}}{1+\frac{1+\cos \left(2\mathrm{x}\right)}{2}}={\int }_0^{\pi }\frac{2\mathrm{dx}}{3+\cos \left(2\mathrm{x}\right)}$$$${=}_{2\mathrm{x}=\mathrm{t}}{\int }_0^{2\pi }\frac{\mathrm{dt}}{3+\mathrm{cost}}{=}_{{\mathrm{e}}^{\mathrm{it}}=\mathrm{z}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(3+\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2})}$$$$=\int \frac{-2\mathrm{idz}}{\mathrm{z}(6+\mathrm{z}+{\mathrm{z}}^{-1})}={\int }_{\mid \mathrm{z}\mid =1}\frac{-2\mathrm{idz}}{{\mathrm{z}}^2+6\mathrm{z}+1}={\int }_{\mid \mathrm{z}\mid =1}\phi \left(\mathrm{z}\right)\mathrm{dz}$$$${\mathrm{\Delta }}^{{}^{\prime }}=9-1=8\Rightarrow {\mathrm{z}}_1=-3+2\sqrt{2}\mathrm{and}{\mathrm{z}}_2=-3-2\sqrt{2}$$$$\mid {\mathrm{z}}_1\mid -1=3-2\sqrt{2}-1=2-2\sqrt{2}<0\Rightarrow \mid {\mathrm{z}}_1\mid <1$$$$\mid {\mathrm{z}}_2\mid -1=3+2\sqrt{2}-1=2+2\sqrt{2}>0\Rightarrow \mid {\mathrm{z}}_2\mid >1\mathrm{residus}\mathrm{theoem}\Rightarrow$$$${\int }_{\mid \mathrm{z}\mid =1}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,{\mathrm{z}}_1)$$$$\phi \left(\mathrm{z}\right)=\frac{-2\mathrm{i}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}\Rightarrow \mathrm{Res}(\phi ,{\mathrm{z}}_1)=\frac{-2\mathrm{i}}{{\mathrm{z}}_1-{\mathrm{z}}_2}=\frac{-2\mathrm{i}}{4\sqrt{2}}=\frac{-\mathrm{i}}{2\sqrt{2}}\Rightarrow$$$${\int }_{\mid \mathrm{z}\mid =1}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{-\mathrm{i}}{2\sqrt{2}}=\frac{\pi }{\sqrt{2}}\Rightarrow \mathrm{\Psi }=\frac{\pi }{\sqrt{2}}$$

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