Question Number 53483 by dwdkswd last updated on 22/Jan/19 Commented by maxmathsup by imad last updated on 22/Jan/19 $${let}\:{A}_{{s}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{s}} }{{e}^{{x}} −\mathrm{1}}\:{dx}\:\Rightarrow{A}_{{s}} =\int_{\mathrm{0}}…
Question Number 119021 by A8;15: last updated on 21/Oct/20 Answered by MJS_new last updated on 21/Oct/20 $$\int\frac{\sqrt{\mathrm{sin}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}}+\sqrt{\mathrm{cos}\:{x}}}{dx}=\int\frac{\sqrt{\mathrm{tan}\:{x}}}{\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)}{dt} \\…
Question Number 184552 by Frix last updated on 08/Jan/23 $${I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{{x}}}\right){dx}=? \\ $$$${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}−\frac{\sqrt{{x}}}{\:\sqrt{\mathrm{2}}}\right){dx}=? \\ $$ Answered…
Question Number 53477 by maxmathsup by imad last updated on 22/Jan/19 $${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\:\sqrt{{x}+{a}}\:+\mathrm{3}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{also}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\:\sqrt{{x}+{a}}\left(\sqrt{{x}+{a}}\:+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\:\sqrt{{x}+\mathrm{1}}+\mathrm{3}}\:\:{and}\:\int_{\mathrm{0}}…
Question Number 53476 by maxmathsup by imad last updated on 22/Jan/19 $${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:{t}\sqrt{\mathrm{2}{t}−\mathrm{1}}{dt}\:\:\:\:{calculate}\:\mid{sup}_{\mathrm{1}\leqslant{x}\leqslant\mathrm{2}} \:{f}\left({x}\right)\:−{inf}_{\mathrm{1}\leqslant{x}\leqslant\mathrm{2}} {f}\left({x}\right)\mid \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 53474 by maxmathsup by imad last updated on 22/Jan/19 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{5}^{\mathrm{2}{x}+\mathrm{1}} \:−\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} }{\mathrm{10}^{{x}} }\:{dx} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 119006 by A8;15: last updated on 21/Oct/20 Answered by Dwaipayan Shikari last updated on 21/Oct/20 $$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} \frac{{tan}^{−\mathrm{1}} {x}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}={I} \\ $$$$=\int_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 53471 by maxmathsup by imad last updated on 22/Jan/19 $$\left.\mathrm{1}\right){find}\:\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{{n}} {tdt}\:\:\:{with}\:{n}\:{integr}\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{U}_{{n}} \\ $$$$…
Question Number 53470 by maxmathsup by imad last updated on 22/Jan/19 $${find}\:\:{Vn}=\int_{\frac{\mathrm{1}}{{n}}} ^{\frac{{an}−\mathrm{1}}{{n}}} \:\frac{\sqrt{{x}}}{\:\sqrt{{a}−\sqrt{{x}}+{x}}}{dx} \\ $$$$ \\ $$ Commented by maxmathsup by imad last updated…
Question Number 53467 by maxmathsup by imad last updated on 22/Jan/19 $${let}\:{A}_{{n}\:{m}} \:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \left(\mathrm{1}−{x}\right)^{{m}} {dx}\:\:{with}\:{n}\:{and}\:{n}\:{integrs}\:{naturals} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}\:{m}} \:\:{by}\:{using}\:{factoriels} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\sum_{{n},{m}} \:{A}_{{nm}} \\ $$…