Question Number 151276 by peter frank last updated on 19/Aug/21 $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)\mathrm{dx} \\ $$ Commented by tabata last updated on 19/Aug/21 $$=\int_{\mathrm{0}} ^{\:\mathrm{2}\pi}…
Question Number 20205 by vivek last updated on 24/Aug/17 $${find}\:{the}\:{sin}^{−\mathrm{1}} \:{diferentiation} \\ $$ Answered by Joel577 last updated on 24/Aug/17 $${y}\:=\:\mathrm{sin}^{−\mathrm{1}} \:\left({x}\right) \\ $$$${x}\:=\:\mathrm{sin}\:{y} \\…
Question Number 151278 by mathdanisur last updated on 19/Aug/21 $$\int\:\frac{\boldsymbol{\mathrm{e}}^{\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}} }{\:\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 19/Aug/21 $$\mathrm{I}=\int\frac{{e}^{\sqrt{{x}−\mathrm{1}}} }{\:\sqrt{{x}−\mathrm{1}}}{dx},\:{u}=\sqrt{{x}−\mathrm{1}}\Rightarrow{du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\:\:=\mathrm{2}\int{e}^{{u}}…
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Question Number 20203 by mondodotto@gmail.com last updated on 24/Aug/17 Commented by Einstein Newton last updated on 24/Aug/17 $$\left(\mathrm{a}\right)\:\mathrm{20}\:+\:\mathrm{37}\:−\:\mathrm{12}\:=\:\mathrm{45} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{20}\:+\:\mathrm{37}\:=\:\mathrm{57} \\ $$ Terms of Service…
Question Number 151272 by Tawa11 last updated on 19/Aug/21 Commented by rs4089 last updated on 19/Aug/21 $${coxeter}\:{integral}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 20202 by Joel577 last updated on 24/Aug/17 $$\mathrm{Is}\:\mathrm{definite}\:\mathrm{integral}\:\mathrm{can}\:\mathrm{have}\:\mathrm{negative}\:\mathrm{value}? \\ $$$$\mathrm{Because}\:\mathrm{I}\:\mathrm{think}\:\int_{{a}} ^{{b}} {f}\left({x}\right)\:{dx}\:\mathrm{is}\:\mathrm{total}\:\mathrm{area}\:\mathrm{below} \\ $$$$\mathrm{graph}\:{f}\left({x}\right)\:\mathrm{from}\:{x}\:=\:{a}\:\mathrm{until}\:{x}\:=\:{b},\:\mathrm{so}\:\mathrm{it}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{be}\:\mathrm{negative} \\ $$ Commented by ajfour last updated…
Question Number 20201 by khamizan833@gmail.con last updated on 24/Aug/17 $$\:^{\mathrm{4}} \sqrt{\mathrm{49}\:−\:\mathrm{20}\sqrt{\mathrm{6}}}\:= \\ $$ Answered by Einstein Newton last updated on 24/Aug/17 $$\sqrt{\mathrm{49}\:−\:\mathrm{20}\sqrt{\mathrm{6}}}\:=\:\sqrt{\mathrm{5}^{\mathrm{2}} \:+\:\left(\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{5}\right)\left(\mathrm{2}\sqrt{\mathrm{6}}\right)} \\…
Question Number 20200 by ketto last updated on 24/Aug/17 Answered by Joel577 last updated on 24/Aug/17 $${x}\:.\:\mathrm{log}\:\mathrm{2}\:=\:\mathrm{6}{x}\:+\:\mathrm{3} \\ $$$$\left({x}\:.\:\mathrm{log}\:\mathrm{2}\right)\:−\:\mathrm{6}{x}\:=\:\mathrm{3} \\ $$$${x}\left[\left(\mathrm{log}\:\mathrm{2}\right)\:−\:\mathrm{6}\right]\:=\:\mathrm{3} \\ $$$${x}\:=\:\frac{\mathrm{3}}{\left(\mathrm{log}\:\mathrm{2}\right)\:−\:\mathrm{6}} \\ $$…
Question Number 151269 by tabata last updated on 19/Aug/21 Commented by tabata last updated on 19/Aug/21 $$?????? \\ $$ Terms of Service Privacy Policy Contact:…