Question Number 129584 by bemath last updated on 16/Jan/21 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{sin}\:\mathrm{x}\right)\left(\mathrm{arctan}\:\mathrm{x}\right)−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}} \right)}=? \\ $$ Answered by liberty last updated on 16/Jan/21 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{\mathrm{x}^{\mathrm{5}}…
Question Number 64047 by aliesam last updated on 12/Jul/19 Commented by mathmax by abdo last updated on 12/Jul/19 $${we}\:{have}\:{x}_{\mathrm{0}} =\mathrm{1}\:{and}\:{x}_{{n}+\mathrm{1}} =\frac{\mathrm{3}+\mathrm{2}{x}_{{n}} }{\mathrm{3}+{x}_{{n}} }\:\Rightarrow{x}_{{n}+\mathrm{1}} ={f}\left({x}_{{n}} \right)\:{with}…
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Question Number 129576 by greg_ed last updated on 16/Jan/21 $$\boldsymbol{\mathrm{please}},\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{f}}\::\:\left[\mathrm{0}\:,\:\boldsymbol{{a}}\right]\:×\:\mathbb{R}_{+} \:\rightarrow\:\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{{x}},\:\boldsymbol{{y}}\right)\:\:\:\:\:\:\: \:\:\boldsymbol{{e}}^{−\boldsymbol{{xy}}} \:\boldsymbol{{sin}}\:\boldsymbol{{x}}\: \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{integrable}}\:???\: \\ $$ Answered by mathmax by…
Question Number 64037 by MJS last updated on 18/Nov/19 $$\mathrm{reduction}\:\mathrm{formulas}\:\mathrm{for}\:{n}\in\mathbb{N},\:\mathrm{some}\:{n}>\mathrm{0},\:\mathrm{some}\:{n}>\mathrm{1} \\ $$$$ \\ $$$$\int\mathrm{sin}^{{n}} \:{x}\:{dx}=−\frac{\mathrm{1}}{{n}}\mathrm{cos}\:{x}\:\mathrm{sin}^{{n}−\mathrm{1}} \:{x}\:+\frac{{n}−\mathrm{1}}{{n}}\int\mathrm{sin}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\int\mathrm{cos}^{{n}} \:{x}\:{dx}=\frac{\mathrm{1}}{{n}}\mathrm{sin}\:{x}\:\mathrm{cos}^{{n}−\mathrm{1}} \:{x}\:+\frac{{n}−\mathrm{1}}{{n}}\int\mathrm{cos}^{{n}−\mathrm{2}} \:{x}\:{dx} \\ $$$$\int\mathrm{tan}^{{n}} \:{x}\:{dx}=\frac{\mathrm{1}}{{n}−\mathrm{1}}\mathrm{tan}^{{n}−\mathrm{1}}…
Question Number 129568 by abdurehime last updated on 16/Jan/21 Commented by greg_ed last updated on 16/Jan/21 $$\boldsymbol{\mathrm{please}},\:\boldsymbol{\mathrm{review}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{question}}\:\mathrm{2}\:! \\ $$$$\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{seems}}\:+\:\boldsymbol{\mathrm{or}}\:−\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{missed}}\:\boldsymbol{\mathrm{before}}\:\mathrm{14}\boldsymbol{{x}}. \\ $$ Commented by bemath last…
Question Number 129569 by zakirullah last updated on 16/Jan/21 $$\mathrm{Qi}.\:\mathrm{using}\:\mathrm{binomial}\:\mathrm{theorem},\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\left(\mathrm{3}^{\mathrm{2n}+\mathrm{2}\:} −\mathrm{8n}−\mathrm{9}\right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{64},\:\mathrm{where} \\ $$$$\:\mathrm{n}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{integer}. \\ $$$$\mathrm{Qii}.\:\:\mathrm{By}\:\mathrm{mathematical}\:\mathrm{induction};\: \\ $$$$\:\:\left(\mathrm{cos}\alpha+\left(\mathrm{n}−\mathrm{1}\right)\beta\right)\:=\:\mathrm{cos}\left(\alpha+\frac{\left(\mathrm{n}−\mathrm{1}\right)\beta}{\mathrm{2}}\right)×\frac{\mathrm{sin}\left(\mathrm{n}\beta/\mathrm{2}\right)}{\mathrm{sin}\left(\beta/\mathrm{2}\right)} \\ $$ Answered by Dwaipayan Shikari…
Question Number 129564 by bramlexs22 last updated on 16/Jan/21 $$\:\mathcal{V}\:=\:\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\left(\mathrm{x}+\theta\right)}\:\mathrm{dx}\: \\ $$ Answered by Lordose last updated on 16/Jan/21 $$\Omega\:=\:\int\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{x}+\theta\right)}\mathrm{dx}\:\overset{\mathrm{u}=\mathrm{x}+\theta} {=}\int\frac{\mathrm{sin}\left(\mathrm{u}−\theta\right)}{\mathrm{sin}\left(\mathrm{u}\right)}\mathrm{du} \\ $$$$\Omega\:=\:\int\frac{\mathrm{sin}\left(\mathrm{u}\right)\mathrm{cos}\theta−\mathrm{cos}\left(\mathrm{u}\right)\mathrm{sin}\theta}{\mathrm{sin}\left(\mathrm{u}\right)}\mathrm{du}\:=\:\mathrm{ucos}\theta\:−\:\mathrm{sin}\theta\mathrm{ln}\left(\mathrm{sinu}\right)\:+\:\mathrm{C} \\ $$$$\Omega\:=\:\left(\mathrm{x}+\theta\right)\mathrm{cos}\theta\:−\:\mathrm{sin}\theta\mathrm{ln}\left(\mathrm{sin}\left(\mathrm{x}+\theta\right)\right)+\:\mathrm{C}…
Question Number 129562 by mathmax by abdo last updated on 16/Jan/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{6}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Lordose last updated on 16/Jan/21…
Question Number 129558 by mnjuly1970 last updated on 16/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{modern}\:\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\:{algebra}\:…\: \\ $$$$\:\:\:\:\:\:\:\::::\:\:{if}\:\:''\:{G}\:''\:{be}\:{a}\:{finite}\:{group}\:{and} \\ $$$$\:\:{O}\:\left({G}\right)={pq}\:\:,\:\:{where}\:''\:{p}\:,\:{q}\:''\:{are}\:{two} \\ $$$$\:\:{prime}\:\:{numbers}\:\left({p}\:>\:{q}\:\right)\:{then}\:{prove}\:{that}: \\ $$$$\:\:{G}\:\:{has}\:\:{at}\:{most}\:{one}\:{subgroup}\:{of}\:{order}\:''\:{p}\:''\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{written}\:{and}\:{compiled}\:{by} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\clubsuit{m}.{n}.{july}.\mathrm{1970}\clubsuit…. \\ $$ Answered…