Question Number 82973 by mathmax by abdo last updated on 26/Feb/20 $${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{x}}{{n}}\right)^{{n}} {arctan}\left({nx}\right){dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148505 by mathdanisur last updated on 28/Jul/21 Answered by dumitrel last updated on 28/Jul/21 $${O}\left(\mathrm{0},\mathrm{0}\right);{A}\left({x},{y}\right);{B}\left(\mathrm{4};\mathrm{3}\right) \\ $$$${OB}=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\mathrm{5} \\ $$$$\Rightarrow{OA}+{AB}={OB}\Rightarrow{A}\in\left({OB}\right)\Rightarrow{x},{y}>\mathrm{0} \\ $$$$\frac{{x}}{\mathrm{4}}=\frac{{y}}{\mathrm{3}}\Rightarrow\mathrm{4}{y}=\mathrm{3}{x}\Rightarrow\mathrm{3}{x}^{\mathrm{2}}…
Question Number 82970 by mathmax by abdo last updated on 26/Feb/20 $$\left.\mathrm{1}\right){find}\:\int\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{9}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{9}} } \\ $$ Commented by mathmax…
Question Number 17435 by 786786AM last updated on 06/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{4}\:\mathrm{sin}\:\frac{\pi}{\mathrm{24}}\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{12}}\:\:\mathrm{cos}\frac{\pi}{\mathrm{6}}. \\ $$ Answered by alex041103 last updated on 07/Jul/17 $$\mathrm{Let}\:{A}=\mathrm{4}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{6}}. \\ $$$$\mathrm{Then}\:\mathrm{we}\:\mathrm{use}\:{sin}\mathrm{2}\theta=\mathrm{2}{sin}\theta{cos}\theta\:: \\ $$$${A}=\mathrm{2}\left(\mathrm{2}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{24}}\right){cos}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{6}}\:\frac{\mathrm{1}}{{cos}\frac{\pi}{\mathrm{24}}} \\…
Question Number 82971 by mathmax by abdo last updated on 26/Feb/20 $$\left.\mathrm{1}\right){find}\:\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{6}} } \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{6}} } \\ $$ Commented by mathmax…
Question Number 148501 by mathmax by abdo last updated on 28/Jul/21 $$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\left\{\mathrm{z}\in\mathrm{C}\:/\mathrm{z}^{\mathrm{n}} \:=\mathrm{1}\right\}\:\:\mathrm{simplify} \\ $$$$\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2n}−\mathrm{1}} \:\mathrm{w}^{\mathrm{p}} \:\:\:\:\:\:\:\:\mathrm{with}\:\mathrm{w}\in\mathrm{U}_{\mathrm{n}} \:\:\: \\ $$$$\mathrm{and}\:\:\sum_{\mathrm{p}=\mathrm{0}} ^{\mathrm{2n}−\mathrm{1}} \left(\mathrm{2w}\:+\mathrm{1}\right)^{\mathrm{p}} \\…
Question Number 148502 by mathmax by abdo last updated on 28/Jul/21 $$\mathrm{let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{2} \\ $$$$\mathrm{simplify}\:\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\:\left(\alpha^{\mathrm{k}} \:+\beta^{\mathrm{k}} \right)\:\:\mathrm{and}\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \left(\:\frac{\mathrm{1}}{\alpha^{\mathrm{k}} }+\frac{\mathrm{1}}{\beta^{\mathrm{k}} }\right) \\ $$…
Question Number 148498 by mathmax by abdo last updated on 28/Jul/21 $$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{2x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 17421 by Tinkutara last updated on 05/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2}^{\mid{x}\mid} \:=\:\mathrm{1}\:+\:\mathrm{2}\mid\mathrm{cos}\:{x}\mid\:\mathrm{is} \\ $$ Answered by mrW1 last updated on 05/Jul/17 $$\mathrm{1}\leqslant\:\mathrm{1}\:+\:\mathrm{2}\mid\mathrm{cos}\:{x}\mid\:\leqslant\mathrm{3} \\ $$$$\mathrm{1}\leqslant\:\mathrm{2}^{\mid\mathrm{x}\mid}…
Question Number 148494 by liberty last updated on 28/Jul/21 $$\:\:\:\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2008}} }{\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{1338}} }\:+\:\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:=\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\:\mathrm{x}=?\: \\ $$ Answered by EDWIN88 last updated on 28/Jul/21 $$\:\mathrm{log}\:_{\mathrm{2}}…