Question Number 218560 by MrGaster last updated on 12/Apr/25 Commented by MrGaster last updated on 12/Apr/25 $${J}_{\mathrm{0}} \left({a}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {a}^{\mathrm{2}{n}} }{\left({n}!\right)^{\mathrm{2}} \mathrm{2}^{\mathrm{2}{n}} }\left(\mathrm{1}−{u}^{\mathrm{2}}…
Question Number 218594 by MrGaster last updated on 12/Apr/25 $$\mathrm{A}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{calculation}\:\mathrm{relatedt} \\ $$$$\mathrm{o}\:\mathrm{arctangent}\:\mathrm{integral}: \\ $$$$\boldsymbol{\mathrm{Exere}}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{\mathrm{2}} }{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}. \\ $$$$\mathrm{Solution}:\underset{{A}} {\underbrace{=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{\mathrm{2}} {x}}{{x}}{dx}}}\:−\underset{{B}} {\underbrace{\int_{\mathrm{0}}…
Question Number 218563 by Nicholas666 last updated on 12/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:{solve}\:{for}\:\boldsymbol{{x}}\:\in\:\mathbb{R}\: \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({xt}\right)}{{e}^{{t}} −\mathrm{1}\:}\:{dt}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\:{coth}\left(\boldsymbol{\pi}{x}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:\: \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 218525 by Spillover last updated on 11/Apr/25 Answered by mr W last updated on 11/Apr/25 Commented by mr W last updated on 11/Apr/25…
Question Number 218526 by Spillover last updated on 11/Apr/25 Commented by mr W last updated on 11/Apr/25 $$\approx\mathrm{0}.\mathrm{8348} \\ $$ Commented by Spillover last updated…
Question Number 218527 by Spillover last updated on 11/Apr/25 Answered by A5T last updated on 12/Apr/25 $$\mathrm{CD}=\mathrm{BD}\:\wedge\:\angle\mathrm{CDB}=\mathrm{90}°\:\Rightarrow\:\mathrm{BC}=\mathrm{CD}\sqrt{\mathrm{2}}=\mathrm{2R} \\ $$$$\Rightarrow\mathrm{CD}=\mathrm{R}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Ptolemy}'\mathrm{s}\:\mathrm{theorem}:\:\mathrm{CD}×\mathrm{AB}+\mathrm{AC}×\mathrm{BD}=\mathrm{AD}×\mathrm{BC} \\ $$$$\Rightarrow\left(\mathrm{AB}+\mathrm{AC}\right)=\mathrm{AD}\sqrt{\mathrm{2}}…\left(\mathrm{i}\right) \\ $$$$\mathrm{AB}^{\mathrm{2}}…
Question Number 218547 by nECxx2 last updated on 11/Apr/25 Commented by nECxx2 last updated on 11/Apr/25 $${Please}\:{help} \\ $$ Answered by MrGaster last updated on…
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Question Number 218543 by agmed last updated on 11/Apr/25 Commented by Ghisom last updated on 11/Apr/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} } \\ $$ Answered by SdC355 last…
Question Number 218539 by SdC355 last updated on 11/Apr/25 $${S}\mathrm{olve} \\ $$$$\frac{\partial^{\mathrm{2}} {w}}{\partial{t}^{\mathrm{2}} }={c}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {w}}{\partial{x}^{\mathrm{2}} } \\ $$$${w}\left(\mathrm{0},{t}\right)={f}\left({t}\right)\:,\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{w}\left({x},{t}\right)=\mathrm{0}\:\left(\mathrm{Boundary}\:\mathrm{Condition}\right) \\ $$$${w}\left({x},\mathrm{0}\right)=\mathrm{0}\:,\:{w}_{{t}} \left({x},\mathrm{0}\right)=\mathrm{0}\:\left(\mathrm{Initial}\:\mathrm{Condition}\right) \\ $$$${f}\left({t}\right)\begin{cases}{\mathrm{sin}\left({t}\right)\:,\:{t}\in\left[\mathrm{0},\mathrm{2}\pi\right)}\\{\mathrm{0}\:,\:\mathrm{otherwise}}\end{cases}…