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Question Number 171560 by mnjuly1970 last updated on 17/Jun/22

$$$$$$\mathrm{Nice}\mathrm{Integral}$$$$$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\frac{tan\left(x\right)}{({cos}^2\left(x\right)+2{sin}^2\left(x\right))}dx=$$

Commented by infinityaction last updated on 17/Jun/22

$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}\frac{\tan x\sec {}^{2}x}{1+2\tan {}^{2}x}dx$$$$\tan x=p\to \sec {}^{2}xdx=dp$$$$\mathrm{\Omega }={\int }_0^1\frac{p}{1+2p^2}dp$$$$1+2p^2=t\to 4pdp=dt$$$$\mathrm{\Omega }=\frac{1}{4}{\int }_1^3\frac{dt}{t}\Rightarrow \frac{1}{4}{\left[\log t\right]}_1^3$$$$\mathrm{\Omega }=\frac{1}{4}\log 3$$

Answered by floor(10²Eta[1]) last updated on 18/Jun/22

$$\mathrm{t}=\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)$$$$\cos \left(\mathrm{x}\right)=1-{2\sin }^2\left(\frac{\mathrm{x}}{2}\right)$$$$1-\frac{2}{{\csc }^2\left(\frac{\mathrm{x}}{2}\right)}=1-\frac{2}{1+{\mathrm{cotg}}^2\left(\frac{\mathrm{x}}{2}\right)}$$$$1-\frac{2}{1+\frac{1}{{\mathrm{t}}^2}}=1-\frac{{2\mathrm{t}}^2}{{\mathrm{t}}^2+1}=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}$$$$\Rightarrow \sin \left(\mathrm{x}\right)=\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}$$$$$$$$\Rightarrow \frac{\mathrm{tgx}}{{\cos }^2\mathrm{x}+{2\sin }^2\mathrm{x}}=\frac{\mathrm{tgx}}{1+{\sin }^2\mathrm{x}}=\frac{\mathrm{tgx}}{2-{\cos }^2\mathrm{x}}$$$$=\frac{\mathrm{sinx}}{2\mathrm{cosx}-{\cos }^3\mathrm{x}}=\frac{\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}}{2\left(\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}\right)-{\left(\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}\right)}^3}$$$$=\frac{2\mathrm{t}{(1+{\mathrm{t}}^2)}^2}{2(1-{\mathrm{t}}^2){(1+{\mathrm{t}}^2)}^2-{(1-{\mathrm{t}}^2)}^3}=\frac{2\mathrm{t}{(1+{\mathrm{t}}^2)}^2}{(1-{\mathrm{t}}^2)[1+{6\mathrm{t}}^2+{\mathrm{t}}^4]}$$$$$$$$\Rightarrow {\int }_0^{\mathrm{tg}\frac{\pi }{8}}\frac{2\mathrm{t}{(1+{\mathrm{t}}^2)}^2}{(1-{\mathrm{t}}^2)({\mathrm{t}}^4+{6\mathrm{t}}^2+1)}\mathrm{dx}$$$${\mathrm{t}}^2=\mathrm{u}$$$$2\mathrm{tdt}=\mathrm{du}$$$${\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{{(1+\mathrm{u})}^2}{(1-\mathrm{u})({\mathrm{u}}^2+6\mathrm{u}+1)}\mathrm{du}$$$$=\frac{1}{2}{\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{du}}{1-\mathrm{u}}-\frac{1}{2}{\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{u}-1}{{\mathrm{u}}^2+6\mathrm{u}+1}\mathrm{du}$$$$=-\frac{1}{2}\ln \mid 1-\mathrm{u}{\mid }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}-\frac{1}{2}{\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{u}-1}{{(\mathrm{u}+3)}^2-8}\mathrm{du}$$$$=-\frac{1}{2}\ln \mid 1-{\mathrm{tg}}^2\frac{\pi }{8}\mid -\frac{1}{2}{\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{u}-1}{(\mathrm{u}+3+2\sqrt{2})(\mathrm{u}+3-2\sqrt{2})}\mathrm{du}$$$$=-\frac{1}{2}\ln \mid 1-{\mathrm{tg}}^2\frac{\pi }{8}\mid -\frac{1}{2}\left(\frac{1+\sqrt{2}}{2}\right){\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{du}}{\mathrm{u}+3+2\sqrt{2}}-\frac{1}{2}\left(\frac{1-\sqrt{2}}{2}\right){\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{du}}{\mathrm{u}+3-2\sqrt{2}}$$$$=-\frac{1}{2}\ln \mid 1-{\mathrm{tg}}^2\frac{\pi }{8}\mid -\frac{1}{2}\left(\frac{1+\sqrt{2}}{2}\right){\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{du}}{\mathrm{u}+3+2\sqrt{2}}-\frac{1}{2}\left(\frac{1-\sqrt{2}}{2}\right){\int }_0^{{\mathrm{tg}}^2\frac{\pi }{8}}\frac{\mathrm{du}}{\mathrm{u}+3-2\sqrt{2}}$$$$=-\frac{1}{2}\ln \mid 1-{\mathrm{tg}}^2\frac{\pi }{8}\mid -\frac{1+\sqrt{2}}{4}{[\ln \mid \mathrm{u}+3+2\sqrt{2}\mid ]}_0^{{\mathrm{tg}}^2\frac{\pi }{8}}-\frac{1-\sqrt{2}}{4}{[\ln \mid \mathrm{u}+3-2\sqrt{2}\mid ]}_0^{{\mathrm{tg}}^2\frac{\pi }{8}}$$$$=-\frac{1}{2}\ln \mid 1-{\mathrm{tg}}^2\frac{\pi }{8}\mid -\frac{1+\sqrt{2}}{4}\ln \mid {\mathrm{tg}}^2\frac{\pi }{8}+3+2\sqrt{2}\mid -\frac{1-\sqrt{2}}{4}\ln \mid {\mathrm{tg}}^2\frac{\pi }{8}+3-2\sqrt{2}\mid +\frac{\sqrt{2}}{2}\ln (3+2\sqrt{2})$$$$$$$$$$

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