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Author: Tinku Tara

For-n-N-n-3-Prove-0-1-2-lt-p-2n-e-2piip-2-e-4piin-d-gt-0-p-is-a-prime-number-

Question Number 221778 by MrGaster last updated on 10/Jun/25 $$\mathrm{For}\:\forall{n}\in\boldsymbol{{N}}^{\ast} ,{n}\geq\mathrm{3}\:\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{\mathrm{2}<{p}\leq\mathrm{2}{n}} {\sum}{e}^{\mathrm{2}\pi{ip}+\alpha} \right)^{\mathrm{2}} {e}^{−\mathrm{4}\pi{in}\alpha} {d}\alpha>\mathrm{0},{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number} \\ $$ Terms of Service Privacy…

0-pi-a-a-cos-2n-x-dx-a-gt-1-0-pi-2-2-cos-4-x-dx-lim-m-0-pi-cos-2n-2mx-a-cos-2n-x-dx-a-gt-1-n-N-

Question Number 221769 by MrGaster last updated on 10/Jun/25 $$\int_{\mathrm{0}} ^{\pi} \frac{{a}}{{a}−\mathrm{cos}^{\mathrm{2}{n}} {x}}{dx}=?,{a}>\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{cos}^{\mathrm{4}} {x}}{dx}=? \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}^{\mathrm{2}{n}} \left(\mathrm{2}{mx}\right)}{{a}−\mathrm{cos}^{\mathrm{2}{n}} {x}}{dx}=?,{a}>\mathrm{1},{n}\in\mathbb{N}^{+}…

Prove-0-1-arcsinx-1-x-4-dx-2-pi-2-16-2-pi-8-ln-2-1-2-n-0-n-2n-1-z-0-2n-1-sin-pi-4-2n-1-

Question Number 221770 by MrGaster last updated on 10/Jun/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arcsin}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\frac{\sqrt{\mathrm{2}}\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{8}}\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\mid{z}_{\mathrm{0}} \mid^{\mathrm{2}{n}+\mathrm{1}} \mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−\left(\mathrm{2}{n}+\mathrm{1}\right)\beta\right) \\ $$ Terms of Service Privacy…

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}}…