Question Number 134678 by mr W last updated on 07/Mar/21 Commented by mr W last updated on 06/Mar/21 $${attempt}\:{to}\:{solve}\:{Q}\mathrm{134376} \\ $$ Commented by mr W…
Question Number 134672 by mnjuly1970 last updated on 06/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:\:\:{calculus}… \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{5}\pi}\leqslant\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{3}}} \frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{2}\pi}\sqrt{\mathrm{8}{sin}\left({x}\right)−{sin}\left(\mathrm{2}{x}\right)}\:{dx}\leqslant\sqrt{\mathrm{6}\pi} \\ $$$$\:\:\:\:\:\:\:\:…{m}.{n}…. \\ $$ Terms of Service Privacy Policy…
Question Number 134668 by EDWIN88 last updated on 06/Mar/21 $$\mathrm{1}+\mathrm{1}+\mathrm{4}−\mathrm{6}−\mathrm{8}−\mathrm{10}+\mathrm{12}+\mathrm{14}+\mathrm{16}−\mathrm{18}−\mathrm{20}−\mathrm{22}+… \\ $$$$\mathrm{S}_{\mathrm{900}} \:=\:? \\ $$ Answered by benjo_mathlover last updated on 06/Mar/21 $$\Rightarrow\underset{\mathrm{6}} {\underbrace{\mathrm{1}+\mathrm{1}+\mathrm{4}}}\:\underset{−\mathrm{24}} {\underbrace{−\mathrm{6}−\mathrm{8}−\mathrm{10}}}\:\underset{\mathrm{42}}…
Question Number 134670 by benjo_mathlover last updated on 06/Mar/21 $$\mid\:\mathrm{x}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\mid\:=\:\sqrt{\mathrm{2}}\:\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\:\right) \\ $$$$\mathrm{find}\:\mathrm{solution} \\ $$ Answered by EDWIN88 last updated on 06/Mar/21 $$\left(\mathrm{1}\right)\:\mathrm{1}−\mathrm{x}^{\mathrm{2}} \:\geqslant\:\mathrm{0}\:\Rightarrow\:−\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{1}…
Question Number 69133 by Henri Boucatchou last updated on 20/Sep/19 $$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int\left(\boldsymbol{\mathrm{x}}\:−\:\boldsymbol{\mathrm{tanx}}\right)^{\mathrm{2}} \:\boldsymbol{\mathrm{dx}}\:\:=\:? \\ $$$$ \\ $$ Commented by mathmax by abdo last updated…
Question Number 134665 by EDWIN88 last updated on 06/Mar/21 $$\mathscr{N}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:+\:\sqrt{\mathrm{x}−\mathrm{1}}}{\:\sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}}\:=? \\ $$ Answered by benjo_mathlover last updated on 06/Mar/21 $$\mathscr{N}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}−\mathrm{1}}\:\left\{\sqrt{\mathrm{x}+\mathrm{1}}+\mathrm{1}\:\right\}}{\:\sqrt{\mathrm{x}−\mathrm{1}}\:\left\{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:\right\}}…
Question Number 69130 by Cmr 237 last updated on 21/Sep/19 $$\mathrm{9}\underset{\mathrm{k}=\mathrm{4}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{3}} −\mathrm{6k}^{\mathrm{2}} +\mathrm{11k}−\mathrm{6}}=? \\ $$$$\:\:\:\:\: \\ $$ Commented by Henri Boucatchou last updated…
Question Number 3595 by Filup last updated on 16/Dec/15 $$\mathrm{I}\:\mathrm{just}\:\mathrm{thought}\:\mathrm{of}\:\mathrm{something}\:\mathrm{I}\:\mathrm{am}\:\mathrm{curious} \\ $$$$\mathrm{in}\:\mathrm{figuring}\:\mathrm{out}. \\ $$$$ \\ $$$$\mathrm{All}\:\mathrm{integer}\:\mathrm{numbers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made}\:\mathrm{up}\:\mathrm{by} \\ $$$${prime}\:{factors}.\:\mathrm{That}\:\mathrm{is}: \\ $$$${n}={p}_{\mathrm{1}} ×{p}_{\mathrm{2}} ×…×{p}_{{i}} \\ $$$${n}\in\mathbb{Z}\:\:\:\:\:\:\:\:\:{p}_{{k}} \in\mathbb{P}…
Question Number 134667 by benjo_mathlover last updated on 06/Mar/21 $$\mathcal{G}\:=\:\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\mathrm{arccos}\:\left(\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{2cos}\:\mathrm{x}}\right)\:\mathrm{dx}\: \\ $$ Commented by LUFFY last updated on 06/Mar/21 $${search}\:\mathrm{C}{oxeter}'{s}\:{integral} \\ $$ Terms…
Question Number 134660 by benjo_mathlover last updated on 06/Mar/21 $$\mathscr{M}\:=\:\int\:\frac{{dx}}{\mathrm{2cos}\:{x}+\mathrm{3sin}\:{x}}\: \\ $$ Answered by EDWIN88 last updated on 06/Mar/21 $$\mathrm{Let}\:\begin{cases}{\mathrm{2}\:=\:\mathrm{r}\:\mathrm{sin}\:\alpha}\\{\mathrm{3}\:=\:\mathrm{r}\:\mathrm{cos}\:\alpha}\end{cases}\:\Rightarrow\:\mathrm{r}^{\mathrm{2}} =\:\mathrm{13};\:\mathrm{r}\:=\sqrt{\mathrm{13}} \\ $$$$\:\mathrm{and}\:\mathrm{tan}\:\alpha\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:,\therefore\:\alpha\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\…