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Question Number 137155 by mnjuly1970 last updated on 30/Mar/21

$$......nicecalculus......$$$$evaluate::$$$$\mathit{\varphi }={\int }_0^{2\pi }\frac{1}{1+{cos}^4\left(x\right)}dx=???$$

Answered by Ar Brandon last updated on 30/Mar/21

$$\varphi ={\int }_0^{2\pi }\frac{\mathrm{dx}}{1+{\cos }^4\mathrm{x}}=4{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^4\mathrm{x}}{{\sec }^4\mathrm{x}+1}\mathrm{dx},\mathrm{t}=\mathrm{tanx}$$$$=4{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2+1}{{({\mathrm{t}}^2+1)}^2+1}\mathrm{dt}=4{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2+1}{{\mathrm{t}}^4+{2\mathrm{t}}^2+2}\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{({\mathrm{t}}^2+\sqrt{2})+({\mathrm{t}}^2-\sqrt{2})}{{\mathrm{t}}^4+{2\mathrm{t}}^2+2}\mathrm{dt}+\frac{2}{\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{({\mathrm{t}}^2+\sqrt{2})-({\mathrm{t}}^2-\sqrt{2})}{{\mathrm{t}}^4+{2\mathrm{t}}^2+2}\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}\left\{\frac{(1+\frac{\sqrt{2}}{{\mathrm{t}}^2})+(1-\frac{\sqrt{2}}{{\mathrm{t}}^2})}{{\mathrm{t}}^2+2+\frac{2}{{\mathrm{t}}^2}}\right\}\mathrm{dt}+\frac{2}{\sqrt{2}}{\int }_0^{\mathrm{\infty }}\left\{\frac{(1+\frac{\sqrt{2}}{{\mathrm{t}}^2})-(1-\frac{\sqrt{2}}{{\mathrm{t}}^2})}{{\mathrm{t}}^2+2+\frac{2}{{\mathrm{t}}^2}}\right\}\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}\{\frac{1+\frac{\sqrt{2}}{{\mathrm{t}}^2}}{{(\mathrm{t}-\frac{\sqrt{2}}{\mathrm{t}})}^2+2\sqrt{2}+2}+\frac{1-\frac{\sqrt{2}}{{\mathrm{t}}^2}}{{(\mathrm{t}+\frac{\sqrt{2}}{\mathrm{t}})}^2+2-2\sqrt{2}}\}\mathrm{dt}$$$$+\frac{2}{\sqrt{2}}{\int }_0^{\mathrm{\infty }}\{\frac{1+\frac{\sqrt{2}}{{\mathrm{t}}^2}}{{(\mathrm{t}-\frac{\sqrt{2}}{\mathrm{t}})}^2+2\sqrt{2}+2}-\frac{1-\frac{\sqrt{2}}{{\mathrm{t}}^2}}{{(\mathrm{t}+\frac{\sqrt{2}}{\mathrm{t}})}^2+2-2\sqrt{2}}\}\mathrm{dt}$$$$=2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{du}}{{\mathrm{u}}^2+2\sqrt{2}+2}+\frac{2}{\sqrt{2}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dv}}{{\mathrm{v}}^2+2\sqrt{2}+2}$$$$={[\frac{2}{\sqrt{2\sqrt{2}+2}}{\tan }^{-1}\left(\frac{\mathrm{u}}{\sqrt{2\sqrt{2}+2}}\right)+\frac{\sqrt{2}}{\sqrt{2\sqrt{2}+2}}{\tan }^{-1}\left(\frac{\mathrm{v}}{\sqrt{2\sqrt{2}+2}}\right)]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}$$$$=\frac{2\pi }{\sqrt{2\sqrt{2}+2}}+\frac{\sqrt{2}\pi }{\sqrt{2\sqrt{2}+2}}$$

Commented by Ar Brandon last updated on 30/Mar/21

$$\mathrm{Not}100\mathrm{\%}\mathrm{sure}...\mathrm{Please}\mathrm{check}$$

Commented by Dwaipayan Shikari last updated on 30/Mar/21

$$\frac{2+\sqrt{2}}{\sqrt{2\sqrt{2}+2}}\pi =\sqrt{1+\sqrt{2}}\pi$$

Commented by mnjuly1970 last updated on 30/Mar/21

$$thanksalotmrbrandon..$$

Commented by mnjuly1970 last updated on 30/Mar/21

$$gratefulmrpayan...$$

Commented by Ar Brandon last updated on 30/Mar/21

Cool

Answered by mathmax by abdo last updated on 31/Mar/21

$$\mathrm{\Phi }={\int }_0^{2\pi }\frac{\mathrm{dx}}{1+{\cos }^4\mathrm{x}}\Rightarrow \mathrm{\Phi }={\int }_0^{2\pi }\frac{\mathrm{dx}}{1+{\left(\frac{1+\cos \left(2\mathrm{x}\right)}{2}\right)}^2}$$$$={\int }_0^{2\pi }\frac{4\mathrm{dx}}{4+(1+2\cos \left(2\mathrm{x}\right)+{\cos }^2\left(2\mathrm{x}\right))}=4{\int }_0^{2\pi }\frac{\mathrm{dx}}{5+2\cos \left(2\mathrm{x}\right)+\frac{1+\cos \left(4\mathrm{x}\right)}{2}}$$$$=8{\int }_0^{2\pi }\frac{\mathrm{dx}}{10+4\cos \left(2\mathrm{x}\right)+1+\cos \left(4\mathrm{x}\right)}=8{\int }_0^{2\pi }\frac{\mathrm{dx}}{11+4\cos \left(2\mathrm{x}\right)+\cos \left(4\mathrm{x}\right)}$$$${=}_{2\mathrm{x}=\mathrm{t}}8{\int }_0^{4\pi }\frac{\mathrm{dt}}{2(11+4\mathrm{cost}+\cos \left(2\mathrm{t}\right))}=4{\int }_0^{4\pi }\frac{\mathrm{dt}}{11+4\mathrm{cost}+{2\cos }^2\mathrm{t}-1}$$$$=4{\int }_0^{4\pi }\frac{\mathrm{dt}}{{2\cos }^2\mathrm{t}+4\mathrm{cost}+10}=2{\int }_0^{4\pi }\frac{\mathrm{dt}}{{\cos }^2\mathrm{t}+2\mathrm{cost}+5}$$$${\mathrm{z}}^2+2\mathrm{z}+5=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1-5=-4\Rightarrow {\mathrm{z}}_1=-1+2\mathrm{i}\mathrm{and}{\mathrm{z}}_2=-1-2\mathrm{i}\Rightarrow$$$${\cos }^2\mathrm{t}+2\mathrm{cost}+5=(\mathrm{cost}-{\mathrm{z}}_1)(\mathrm{cost}-{\mathrm{z}}_2)\Rightarrow$$$$\frac{1}{{\cos }^2\mathrm{t}+2\mathrm{cost}+5}=\frac{1}{(\mathrm{cost}-{\mathrm{z}}_1)(\mathrm{cost}-{\mathrm{z}}_2)}$$$$=(\frac{1}{\mathrm{cost}-{\mathrm{z}}_1}-\frac{1}{\mathrm{cost}-{\mathrm{z}}_2})\times \frac{1}{4\mathrm{i}}\Rightarrow$$$$\mathrm{\Phi }=\frac{1}{2\mathrm{i}}{\int }_0^{4\pi }\frac{\mathrm{dt}}{\mathrm{cost}-{\mathrm{z}}_1}-\frac{1}{2\mathrm{i}}{\int }_0^{4\pi }\frac{\mathrm{dt}}{\mathrm{cost}-{\mathrm{z}}_2}$$$$=-\mathrm{i}{\int }_0^{2\pi }\frac{\mathrm{dt}}{\mathrm{cost}-{\mathrm{z}}_1}+\mathrm{i}{\int }_0^{2\pi }\frac{\mathrm{dt}}{\mathrm{cost}-{\mathrm{z}}_2}=-\mathrm{iH}+\mathrm{iK}$$$$\mathrm{H}{=}_{\mathrm{z}={\mathrm{e}}^{\mathrm{it}}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(\frac{\mathrm{z}+{\mathrm{z}}^{-1}}{2}+1-2\mathrm{i})}$$$$={\int }_{\mid \mathrm{z}\mid =1}\frac{2\mathrm{dz}}{\mathrm{iz}(\mathrm{z}+{\mathrm{z}}^{-1}+2-4\mathrm{i})}=\int \frac{-2\mathrm{i}}{{\mathrm{z}}^2+1+(2-4\mathrm{i})\mathrm{z}}$$$$\phi \left(\mathrm{z}\right)=\frac{-2\mathrm{i}}{{\mathrm{z}}^2+(2-4\mathrm{i})\mathrm{z}+1}\mathrm{poles}?$$$${\mathrm{\Delta }}^{{}^{\prime }}={(1-2\mathrm{i})}^2-1=1-4\mathrm{i}-4-1=-4(1+\mathrm{i})=-4\sqrt{2}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\Rightarrow$$$$\sqrt{{\mathrm{\Delta }}^{,}}=2\mathrm{i}\left({}^4\sqrt{2}\right){\mathrm{e}}^{\frac{\mathrm{i}\pi }{8}}\Rightarrow {\mathrm{z}}_1=-1+2\mathrm{i}+2\mathrm{i}\left({}^4\sqrt{2}\right){\mathrm{e}}^{\frac{\mathrm{i}\pi }{8}}$$$$=-1+2\mathrm{i}+2\mathrm{i}\left({}^4\sqrt{2}\right)(\frac{\sqrt{2+\sqrt{2}}}{2}+\mathrm{i}\frac{\sqrt{2-\sqrt{2}}}{2})$$$$=-1+2\mathrm{i}+2\mathrm{i}\left({}^4\sqrt{2}\right)\frac{\sqrt{2+\sqrt{2}}}{2}-\left({}^4\sqrt{2}\right)\sqrt{2-\sqrt{2}}$$$$=-1+2\mathrm{i}+\mathrm{i}\left({}^4\sqrt{2}\right)\sqrt{2+\sqrt{2}}-\left({}^4\sqrt{2}\right)\sqrt{2-\sqrt{2}}$$$$\Rightarrow \mid {\mathrm{z}}_1\mid =\sqrt{{(1+\left({}^4\sqrt{2}\right)\sqrt{2-\sqrt{2}})}^2+\sqrt{2}(2+\sqrt{2})}....\mathrm{be}\mathrm{continued}....$$

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