Question Number 61667 by aliesam last updated on 06/Jun/19 $$\int\sqrt{{tan}\left({x}\right)}\:{dx}\: \\ $$ Answered by MJS last updated on 06/Jun/19 $$\int\sqrt{\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}}…
Question Number 61662 by maxmathsup by imad last updated on 06/Jun/19 $${calculate}\:\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{cosx}}{{e}^{\frac{\mathrm{1}}{{x}}} \:+\mathrm{1}}\:{dx}\: \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 61661 by maxmathsup by imad last updated on 05/Jun/19 $$\left.\mathrm{1}\right)\:{calculate}\:\int\int_{{R}^{+^{\mathrm{2}} } } \:\:\:\:\:\frac{{dxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:{dx}\:. \\ $$ Commented…
Question Number 61660 by maxmathsup by imad last updated on 05/Jun/19 $${let}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}} }\:{dt}\:\:\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} } \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{serie}\:\Sigma{ln}\left(\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\right)\:\:{and}\:{prove}\:\:{that}\:{lim}_{{n}\rightarrow+\infty}…
Question Number 127190 by bramlexs22 last updated on 27/Dec/20 $$\:\int\:\frac{\sqrt{{a}}−\sqrt{{x}}}{\mathrm{1}−\sqrt{{ax}}}\:{dx}\:=?\:;\:{a}>\mathrm{0} \\ $$ Answered by Dwaipayan Shikari last updated on 27/Dec/20 $$\Rightarrow{ax}={u}^{\mathrm{2}} \Rightarrow{a}=\mathrm{2}{u}\frac{{du}}{{dx}} \\ $$$$\mathrm{2}\int\frac{\sqrt{{a}}−\frac{{u}}{\:\sqrt{{a}}}}{\mathrm{1}−{u}}.\frac{{u}}{{a}}{du}\:=\:\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{3}} }}\int\frac{{a}−{u}}{\mathrm{1}−{u}}{du}=\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{3}}…
Question Number 61654 by aliesam last updated on 05/Jun/19 $$\int_{\mathrm{0}} ^{\infty} {e}^{−{e}^{{x}} } {ln}\left({x}\right)\:{dx}\:=\:\mathrm{0}.\mathrm{27634} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61648 by maxmathsup by imad last updated on 05/Jun/19 $${calculate}\:\int\int_{{W}} \:\left({x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}}{dxdy}\:\:\:\:{with} \\ $$$${W}\:=\left\{\:\left({x},{y}\right)\:\in\:{R}^{\mathrm{2}} \:\:/\:\:\:\:\mathrm{1}\leqslant{x}\:\leqslant\sqrt{\mathrm{3}}\:\:{and}\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{2}{y}\:\leqslant\:\mathrm{2}\:\right\} \\ $$ Commented…
Question Number 61645 by maxmathsup by imad last updated on 05/Jun/19 $${calculate}\:\int\int_{{D}} \int\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dxdydz} \\ $$$${with}\:{D}\:=\left\{\left({x},{y},{z}\right)\:/\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:,\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\:\:,\:\mathrm{2}\leqslant{z}\leqslant\mathrm{3}\:\right\} \\ $$ Commented by maxmathsup by imad…
Question Number 192715 by mustafazaheen last updated on 25/May/23 $$\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }}=? \\ $$ Answered by Frix last updated on 25/May/23 $$\int\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}}…
Question Number 192712 by cortano12 last updated on 25/May/23 $$\:\:\:\:\underset{−\mathrm{1}/\mathrm{2}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{dx}\:=? \\ $$ Answered by horsebrand11 last updated on 25/May/23 $${I}=\mathrm{2}\underset{\mathrm{0}}…