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Author: Tinku Tara

Question-218779

Question Number 218779 by Spillover last updated on 15/Apr/25 Answered by breniam last updated on 15/Apr/25 $$ \\ $$$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosh}\:{x}}\mathrm{d}{x}=\int\frac{\mathrm{1}}{\mathrm{1}+\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}}\mathrm{d}{x}=\int\frac{\mathrm{2}}{{e}^{{x}} +\mathrm{2}+{e}^{−{x}} }\mathrm{d}{x}= \\ $$$$\int\frac{\mathrm{2}}{\left({e}^{\frac{{x}}{\mathrm{2}}}…

Question-218775

Question Number 218775 by sonukgindia last updated on 15/Apr/25 Answered by SdC355 last updated on 15/Apr/25 $$\sqrt{{x}+\mathrm{22}}\in\mathbb{Z} \\ $$$$\mathrm{1}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{8}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{27}^{\mathrm{1}/\mathrm{3}} \:,\:\mathrm{64}^{\mathrm{1}/\mathrm{3}} …\mathrm{etc} \\ $$$${x}=−\mathrm{21}\:,\:−\mathrm{14}\:,\:\mathrm{5}\:,\:….…

Fourier-Series-f-e-izsin-

Question Number 218769 by SdC355 last updated on 15/Apr/25 $$\mathrm{Fourier}\:\mathrm{Series}\:{f}\left(\theta\right)={e}^{\boldsymbol{{i}}{z}\mathrm{sin}\left(\theta\right)} \\ $$ Answered by MrGaster last updated on 16/Apr/25 $${e}^{{iz}\:\mathrm{sin}\left(\theta\right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({iz}\:\mathrm{sin}\left(\theta\right)\right)^{{n}} }{{n}!} \\…

0-J-t-J-t-dt-0-

Question Number 218771 by SdC355 last updated on 15/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left(\alpha{t}\right){J}_{\nu} \left(\beta{t}\right)\mathrm{d}{t}=?? \\ $$$$\left(\alpha,\beta\neq\mathrm{0}\right) \\ $$ Answered by Nicholas666 last updated on 16/Apr/25…

J-z-

Question Number 218767 by SdC355 last updated on 15/Apr/25 $$\underset{\ell\in\left(−\infty,\infty\right)} {\sum}\:{J}_{\ell} \left({z}\right)=?? \\ $$ Answered by MrGaster last updated on 15/Apr/25 $$\underset{\ell\in\left(−\infty,\infty\right)} {\sum}{J}_{\ell} \left({z}\right)=\mathrm{exp}\left(\frac{{z}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}}\right)\right)=\mathrm{exp}\left(\mathrm{0}\right)=\mathrm{1} \\…

Question-218676

Question Number 218676 by Spillover last updated on 14/Apr/25 Answered by mr W last updated on 14/Apr/25 $$\mathrm{cos}\:\theta=\frac{{R}−{r}}{{R}+{r}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\sqrt{\frac{{r}}{{R}+{r}}} \\ $$$${x}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\mathrm{2}{R}\sqrt{\frac{{r}}{{R}+{r}}}\:\:\checkmark \\ $$…