Question Number 151851 by Tawa11 last updated on 23/Aug/21 Answered by OlafThorendsen last updated on 23/Aug/21 $$\mathrm{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}} \left({x}^{\mathrm{2}} +\mathrm{ln}\left(\frac{\pi+{x}}{\pi−{x}}\right)\right)\mathrm{cos}{x}\:{dx}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{u}\:=\:−{x}\:: \\ $$$$\mathrm{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}}…
Question Number 20777 by Tinkutara last updated on 02/Sep/17 $${In}\:{the}\:{figure}\:{shown},\:{mass}\:'{m}'\:{is} \\ $$$${placed}\:{on}\:{the}\:{inclined}\:{surface}\:{of}\:{a} \\ $$$${wedge}\:{of}\:{mass}\:{M}.\:{All}\:{the}\:{surfaces} \\ $$$${are}\:{smooth}.\:{Find}\:{the}\:{acceleration}\:{of} \\ $$$${the}\:{wedge}. \\ $$ Commented by Tinkutara last updated…
Question Number 86313 by john santu last updated on 28/Mar/20 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}} \\ $$ Commented by john santu last updated on 28/Mar/20 $$\frac{\sqrt{\mathrm{sin}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:+\:\frac{\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:=\:\mathrm{1} \\…
Question Number 151841 by mathdanisur last updated on 23/Aug/21 $$\mathrm{The}\:\mathrm{volue}\:\mathrm{of}\:\mathrm{the}\:\mathrm{limit}:\: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{−\boldsymbol{\mathrm{n}}^{\mathrm{2}} } }{\underset{\boldsymbol{\mathrm{k}}=\boldsymbol{\mathrm{n}}+\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{−\boldsymbol{\mathrm{k}}^{\mathrm{2}} } }\:\:\:;\:\:\:\left(\mathrm{a}\right)\mathrm{0}\:\:\left(\mathrm{b}\right)\mathrm{some}\:\mathrm{c}\in\left(\mathrm{0};\mathrm{1}\right)\:\:\left(\mathrm{c}\right)\mathrm{1} \\ $$ Terms of Service Privacy…
Question Number 86307 by niroj last updated on 28/Mar/20 $$\:\:\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\mathrm{2}\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:+\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{x}}. \\ $$ Answered by TANMAY PANACEA. last updated on 28/Mar/20…
Question Number 86302 by john santu last updated on 28/Mar/20 $$\int\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}\:=? \\ $$ Commented by abdomathmax last updated on 28/Mar/20 $${I}\:=\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}\:{changement}\:{x}\:=\mathrm{2}{sh}\left({t}\right)\:{give}…
Question Number 151838 by talminator2856791 last updated on 23/Aug/21 $$\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\left({x}^{\mathrm{log}\left(\lfloor\left(\lfloor{x}\rfloor!\right)^{\left(\mathrm{log}\left(\lfloor{x}−\mathrm{1}\rfloor!\right)\right)^{−\mathrm{1}} } \rfloor\right)+\mathrm{1}} +\mathrm{1}\right)^{{x}} }{\lfloor{x}^{\mathrm{log}\left({x}^{{x}} \right)+\mathrm{1}} \rfloor!+\mathrm{1}}\:{dx} \\ $$$$\: \\ $$ Terms…
Question Number 20764 by Tinkutara last updated on 02/Sep/17 $${A}\:{small}\:{bead}\:{is}\:{slipped}\:{on}\:{a}\:{horizontal} \\ $$$${rod}\:{of}\:{length}\:{l}.\:{The}\:{rod}\:{starts}\:{moving} \\ $$$${with}\:{a}\:{horizontal}\:{acceleration}\:{a}\:{in}\:{a} \\ $$$${direction}\:{making}\:{an}\:{angle}\:\alpha\:{with}\:{the} \\ $$$${length}\:{of}\:{the}\:{rod}.\:{Assuming}\:{that} \\ $$$${initially}\:{the}\:{bead}\:{is}\:{in}\:{the}\:{middle}\:{of} \\ $$$${the}\:{rod},\:{find}\:{the}\:{time}\:{elapsed}\:{before} \\ $$$${the}\:{bead}\:{leaves}\:{the}\:{rod}.\:{Coefficient}\:{of} \\…
Question Number 86298 by 21042009 last updated on 28/Mar/20 $${let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{2} \\ $$$${x}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}=\pm\sqrt{\mathrm{2}} \\ $$$${then}\:{let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{4} \\ $$$${x}^{\mathrm{4}}…
Question Number 151832 by DELETED last updated on 23/Aug/21 Answered by DELETED last updated on 23/Aug/21 $$\left.\mathrm{57}\right).\:\mathrm{R}_{\mathrm{s}} =\mathrm{2R}+\mathrm{2R}=\mathrm{4R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }=\frac{\mathrm{1}}{\mathrm{2R}}+\frac{\mathrm{1}}{\mathrm{4R}}=\frac{\mathrm{3}}{\mathrm{4R}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\mathrm{p}} =\frac{\mathrm{4R}}{\mathrm{3}}\: \\…