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Author: Tinku Tara

Prove-0-1-k-1-1-x-k-dx-4pi-3-23-sinh-23-pi-6-2-cosh-23-pi-3-1-

Question Number 221663 by MrGaster last updated on 09/Jun/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{x}^{{k}} \right){dx}=\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{23}}}\centerdot\frac{\mathrm{sinh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{6}}}{\mathrm{2}\:\mathrm{cosh}\frac{\sqrt{\mathrm{23}}\pi}{\mathrm{3}}−\mathrm{1}} \\ $$ Commented by MrGaster last updated on 09/Jun/25 It is difficult for me to give an analytical solution to that integral.…

Question-221721

Question Number 221721 by ahmed2025 last updated on 09/Jun/25 Answered by wewji12 last updated on 09/Jun/25 $$\frac{\mathrm{d}{y}}{\mathrm{dln}\left({t}\right)}=\frac{\frac{\mathrm{d}{y}}{\mathrm{d}{t}}}{\frac{\mathrm{d}\left\{\mathrm{ln}\left({t}\right)\right\}}{\mathrm{d}{t}}}\:\left(\mathrm{Warning}\:\:\frac{\mathrm{d}{y}}{\mathrm{d}{x}}\:\mathrm{is}\:\mathrm{NOT}\:\mathrm{Fraction}!!!\right) \\ $$$$\frac{\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left\{{t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)\right\}}{\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left\{\mathrm{ln}\left({t}\right)\right\}}=\frac{\mathrm{2}{t}\mathrm{ln}\left({t}\right)+{t}}{\frac{\mathrm{1}}{{t}}}={t}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)={t}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2ln}\left({t}\right)\right) \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}\left\{\mathrm{ln}\left({t}\right)\right\}}={t}^{\mathrm{2}}…

81-1-64-1-27-1-3-4-0-4-3-4-

Question Number 221754 by fantastic last updated on 09/Jun/25 $$\left.\sqrt[{\sqrt[{\sqrt[{\:\mathrm{3}^{\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } } }]{\mathrm{27}}}]{\mathrm{64}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$ Answered by wewji12 last updated on 09/Jun/25 $$…..\mathrm{what}\:\mathrm{a}\:\mathrm{horrible}\:\mathrm{notation} \\…

Question-221686

Question Number 221686 by Tawa11 last updated on 09/Jun/25 Commented by AlagaIbile last updated on 09/Jun/25 $$\:{Let}\:\mathrm{tan}\:\boldsymbol{{x}}\:=\:\boldsymbol{{y}},\:\mathrm{cos}\:\boldsymbol{{x}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}} \\ $$$$\Rightarrow\:\mathrm{cos}^{\mathrm{2}\:} \left[\mathrm{cos}^{-\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}}\right] \\ $$$$\Rightarrow\:\left[\mathrm{cos}\:\left(\mathrm{cos}^{-\mathrm{1}}…