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Question Number 77160 by Chi Mes Try last updated on 03/Jan/20

Commented by Chi Mes Try last updated on 03/Jan/20

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Answered by mind is power last updated on 05/Jan/20

$${\int }_0^{\pi }\frac{\mathrm{xsin}\left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}={\int }_0^{\pi }\frac{(\pi -\mathrm{x})\sin \left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$$$$\Rightarrow 2\int \frac{\mathrm{xsin}\left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}=\pi {\int }_0^{\pi }\frac{\sin \left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}={[-\pi \arctan \left(\cos \left(\mathrm{x}\right)\right)]}_0^{\pi }$$$$=\frac{{\pi }^2}{2}\Rightarrow {\int }_0^{\pi }\frac{\mathrm{xsin}\left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}=\frac{{\pi }^2}{4}$$$${\left(\mathrm{k}\left(\frac{1}{\sqrt{2}}\right)\right)}^2={\left({\int }_0^{\frac{\pi }{2}}\frac{\sqrt{2}\mathrm{dx}}{\sqrt{2-{\sin }^2\left(\mathrm{x}\right)}}\right)}^2=2{\left({\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\sqrt{2-{\sin }^2\left(\mathrm{x}\right)}}\right)}^2$$$$\mathrm{\Gamma }\left(\frac{1}{4}\right).\mathrm{\Gamma }(1-\frac{1}{4})=\frac{\pi }{\sin \left(\frac{\pi }{4}\right)}=\pi \sqrt{2}$$$$\frac{\mathrm{\Gamma }\left(\mathrm{x}\right).\mathrm{\Gamma }\left(\mathrm{y}\right)}{\mathrm{\Gamma }(\mathrm{x}+\mathrm{y})}={\int }_0^1{\mathrm{t}}^{\mathrm{x}-1}{(1-\mathrm{t})}^{\mathrm{y}-1}\mathrm{dt}$$$$2\mathrm{x}=\mathrm{y}=\frac{1}{2}\Rightarrow$$$$\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}={\int }_0^1{\mathrm{t}}^{\frac{-3}{4}}.{(1-\mathrm{t})}^{-\frac{1}{2}}\mathrm{dt}$$$$\mathrm{t}={\mathrm{u}}^4\Rightarrow \mathrm{du}=\frac{1}{4}{\mathrm{t}}^{-\frac{3}{4}}\mathrm{dt}$$$$\Rightarrow \frac{\mathrm{\Gamma }\left(\frac{1}{4}\right).\sqrt{\pi }}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}=4{\int }_0^1{(1-{\mathrm{u}}^4)}^{-\frac{1}{4}}\mathrm{du}$$$$\mathrm{\Gamma }\left(\frac{1}{4}\right).\mathrm{\Gamma }\left(\frac{3}{4}\right)=\pi \sqrt{2}$$$$\Rightarrow \frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\sqrt{2\pi }}=4{\int }_0^1\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^4}}$$$$\mathrm{k}{\left(\frac{1}{\sqrt{2}}\right)}^2=2{\left({\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\sqrt{2-{\sin }^2\left(\mathrm{x}\right)}}\right)}^2$$$$=2{\left({\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\sqrt{1+{\cos }^2\left(\mathrm{x}\right)}}\right)}^2$$$$=2{\left({\int }_0^1\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^4}}\right)}^2=\mathrm{k}{\left(\frac{1}{\sqrt{2}}\right)}^2\Rightarrow \frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{4\sqrt{2\pi }}={\int }_0^1\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^4}}$$$$\Rightarrow 2{\left({\int }_0^1\left(\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^4}}\right)\right)}^2={\mathrm{k}}^2\left(\frac{1}{\sqrt{2}}\right)=\frac{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}{16\pi }=\frac{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}{4\pi }$$$$\frac{{\mathrm{k}}^2\left(\frac{1}{\sqrt{2}}\right)}{{\int }_0^{\pi }\frac{\mathrm{xsin}\left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}}=\frac{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}{\frac{16\pi }{\frac{{\pi }^2}{4}}}=\frac{{\mathrm{\Gamma }}^4\left(\frac{1}{4}\right)}{4{\pi }^3}$$$$$$$$$$$$$$$$$$$$$$

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