Question Number 148564 by mathmax by abdo last updated on 29/Jul/21 $$\mathrm{calculate}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{logx}}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$ Answered by Kamel last updated on 29/Jul/21 $$ \\…
Question Number 148567 by mathmax by abdo last updated on 29/Jul/21 $$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{logx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dx} \\ $$ Answered by Ar Brandon last updated…
Question Number 83028 by jagoll last updated on 27/Feb/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{y}\:\mathrm{sec}\:\mathrm{x}\:=\:\mathrm{tan}\:\mathrm{x} \\ $$ Commented by john santu last updated on 27/Feb/20 $$\mathrm{IF}\:\Rightarrow\:\mathrm{e}^{\int\:\mathrm{sec}\:\mathrm{x}\:\mathrm{dx}} \:=\:\mathrm{e}\:^{\mathrm{ln}\:\left(\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}\:\right)\:} \\ $$$$\mathrm{IF}\:=\:\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}\: \\…
Question Number 17493 by Arnab Maiti last updated on 06/Jul/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{logarithm}\:\mathrm{series}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148566 by mathmax by abdo last updated on 29/Jul/21 $$\mathrm{calculate}\:\:\mathrm{U}_{\mathrm{n}} =\int\int_{\left[\frac{\mathrm{1}}{\mathrm{n}},\mathrm{n}\left[\right.\right.} \:\:\:\frac{\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} }\mathrm{dxdy} \\ $$$$\mathrm{and}\:\mathrm{determine}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} \\ $$$$\mathrm{nature}\:\mathrm{of}\:\Sigma\:\mathrm{U}_{\mathrm{n}} ? \\…
Question Number 17492 by Tinkutara last updated on 06/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{lying}\:\mathrm{in} \\ $$$$\left[−\pi,\:\pi\right]\:\mathrm{and}\:\mathrm{satisfying}\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:=\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\mathrm{and}\:\mathrm{sin}\:\mathrm{2}\theta\:+\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:−\:\mathrm{cos}\:\theta\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{is} \\ $$ Answered by ajfour last updated on 08/Jul/17 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2sin}\:^{\mathrm{2}}…
Question Number 83026 by naka3546 last updated on 27/Feb/20 $${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\mathrm{123456}..\mathrm{201820192020}\:\:{divided}\:\:{by}\:\:\mathrm{13}\:\:,\:\:{the} \\ $$$${remainder}\:\:{is}\:\:\mathrm{5}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148559 by Jonathanwaweh last updated on 29/Jul/21 Answered by Kamel last updated on 29/Jul/21 $${a}=\mathrm{3}{k}+{r},{b}=\mathrm{3}{k}'+{r}'\:\mathrm{0}\leqslant{r}<\mathrm{3},\:\mathrm{0}\leqslant{r}'<\mathrm{3}. \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{3}{c}=\mathrm{9}\left({k}^{\mathrm{2}} +{k}'^{\mathrm{2}} \right)+\mathrm{6}\left({kr}+{k}'{r}'\right)+{r}^{\mathrm{2}} +{r}'^{\mathrm{2}} \\…
Question Number 83020 by mathmax by abdo last updated on 27/Feb/20 $${find}\:{the}\:{sequence}\:{u}_{{n}} \:{wich}\:{verify}\:{u}_{{n}} +{u}_{{n}+\mathrm{1}} =\frac{{sin}\left({n}\right)}{{n}}\:\:\forall{n}>\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148558 by puissant last updated on 29/Jul/21 $$\mathrm{Trouver}\:\mathrm{toutes}\:\mathrm{les}\:\mathrm{fonctions}\:\mathrm{continues} \\ $$$$\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{verifiant}: \\ $$$$\forall\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{R}^{\mathrm{2}} ,\:\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{f}\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{f}^{\mathrm{2}} \left(\mathrm{y}\right).. \\ $$$$\mathrm{monsieur}\:\mathrm{j}'\mathrm{ai}\:\mathrm{suppos}\acute {\mathrm{e}}\:\mathrm{que}\:\mathrm{f}\:\mathrm{est}\:\mathrm{un}\: \\ $$$$\mathrm{morphisme}\:\mathrm{mutiplicatif}\:\mathrm{de}\:\mathbb{R}..\:\mathrm{mais}\:\mathrm{ca}\:\mathrm{ne} \\ $$$$\mathrm{sort}\:\mathrm{pas}… \\…