Question Number 38105 by maxmathsup by imad last updated on 21/Jun/18 $${find}\:\int\:\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}\:+\sqrt{\mathrm{2}{x}−\mathrm{1}}}\: \\ $$ Answered by MJS last updated on 22/Jun/18 $$\int\frac{{dx}}{\:\sqrt{\mathrm{2}{x}+\mathrm{1}}+\sqrt{\mathrm{2}{x}−\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\sqrt{\mathrm{2}{x}+\mathrm{1}}−\sqrt{\mathrm{2}{x}−\mathrm{1}}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{2}{x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 38102 by maxmathsup by imad last updated on 21/Jun/18 $${let}\:{B}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } {dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{B}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{B}_{{n}} \\ $$ Terms…
Question Number 38103 by maxmathsup by imad last updated on 21/Jun/18 $${find}\:{I}\left(\lambda\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{xdx}}{\lambda\:+{tanx}}\:\:\lambda\:{from}\:{R}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 38100 by maxmathsup by imad last updated on 21/Jun/18 $${let}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \left({x}−\left[{x}\right]\right)^{\mathrm{2}} {dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$ Commented by…
Question Number 38101 by maxmathsup by imad last updated on 21/Jun/18 $${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:{e}^{{x}−\left[{x}\right]} {dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$ Commented by…
Question Number 169169 by infinityaction last updated on 25/Apr/22 Commented by infinityaction last updated on 25/Apr/22 $$\left(\mathrm{0731619}\right)\:{your}\:{old}\:{question}\:{reposted} \\ $$ Commented by infinityaction last updated on…
Question Number 38074 by ajfour last updated on 21/Jun/18 $$\int\frac{{dx}}{{a}+{b}\mathrm{tan}\:^{\mathrm{2}} {x}}\:=\:? \\ $$ Commented by prof Abdo imad last updated on 21/Jun/18 $${changement}\:{tanx}={t}\:{give} \\ $$$${I}\:=\:\int\:\:\frac{\mathrm{1}}{{a}+{bt}^{\mathrm{2}}…
Question Number 103607 by bemath last updated on 16/Jul/20 $${what}\:{is}\:{the}\:{value}\:{of}\:\int_{{c}} \left({x}+\mathrm{2}{y}\right){dx}+\left(\mathrm{4}−\mathrm{2}{x}\right){dy} \\ $$$${around}\:{the}\:{ellipse}\:{C}:\:\frac{{x}^{\mathrm{2}} }{\mathrm{16}}+\frac{{y}^{\mathrm{2}} }{\mathrm{8}}=\mathrm{1} \\ $$$${in}\:{the}\:{counterclockwise} \\ $$$${direction}\:?\: \\ $$ Answered by bobhans last…
Question Number 38057 by ajfour last updated on 21/Jun/18 $$\int\frac{\mathrm{cos}\:\mathrm{5}{x}+\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{1}−\mathrm{2cos}\:\mathrm{3}{x}}{dx}\:\:=\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jun/18 $$\int\frac{{sin}\mathrm{3}{x}\left({cos}\mathrm{5}{x}+{coz}\mathrm{4}{x}\right)}{{sin}\mathrm{3}{x}−{sin}\mathrm{6}{x}}{dx} \\ $$$$\int\frac{{sin}\mathrm{3}{x}\left({cos}\mathrm{5}{x}+{cos}\mathrm{4}{x}\right)}{−\mathrm{2}{cos}\frac{\mathrm{9}{x}}{\mathrm{2}}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{dx} \\ $$$$\int\frac{\mathrm{2}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}×\mathrm{2}{cos}\frac{\mathrm{9}{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{−\mathrm{2}{cos}\frac{\mathrm{9}{x}}{\mathrm{2}}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{dx} \\…
Question Number 38058 by ajfour last updated on 21/Jun/18 $$\int\frac{\mathrm{tan}\:{x}}{{a}+{b}\mathrm{tan}\:^{\mathrm{2}} {x}}\:{dx}\:\:=\:? \\ $$ Answered by behi83417@gmail.com last updated on 21/Jun/18 $${tg}^{\mathrm{2}} {x}=\frac{{a}}{{b}}{t}\Rightarrow\mathrm{2}{tgx}\left(\mathrm{1}+{tg}^{\mathrm{2}} {x}\right){dx}=\frac{{a}}{{b}}{dt} \\ $$$$\mathrm{2}{tgxdx}=\frac{\frac{{a}}{{b}}{dt}}{\mathrm{1}+\frac{{a}}{{b}}{t}}\Rightarrow{tgxdx}=\frac{{adt}}{\mathrm{2}\left({b}+{at}\right)}…