Question Number 29202 by ajfour last updated on 05/Feb/18 $${Find}\:{area}\:{between}\:{by}\:{y}=\mathrm{1}\:\:{and} \\ $$$${y}=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:. \\ $$ Answered by mrW2 last updated on 05/Feb/18 $${y}=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 94735 by student work last updated on 20/May/20 $$\int\frac{\mathrm{2t}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)\left(\mathrm{1}+\mathrm{t}\right)}\mathrm{dt}=? \\ $$ Commented by student work last updated on 20/May/20 $$\mathrm{who}\:\mathrm{can}\:\mathrm{another}? \\ $$…
Question Number 94718 by john santu last updated on 20/May/20 $$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:= \\ $$$$\int\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx}\:+\:\int\:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{1}−\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }}\:\mathrm{dx}\:+\: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{dt}}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\int\:\frac{−\mathrm{du}}{\:\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}}…
Question Number 94701 by i jagooll last updated on 20/May/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{2} \\ $$$$ \\ $$ Commented by mr W last updated on…
Question Number 29162 by abdo imad last updated on 04/Feb/18 $${find}\:\:{find}\:{I}=\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\:\frac{\mid{x}−\mathrm{2}\mid}{\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} }{dx}\:. \\ $$ Commented by abdo imad last updated on 06/Feb/18…
Question Number 160204 by amin96 last updated on 25/Nov/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{3}} }=? \\ $$ Commented by mr W last updated on 25/Nov/21 $${i}\:{think}\:\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 94662 by i jagooll last updated on 20/May/20 $$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}+\mathrm{cot}\:\mathrm{x}}\:\mathrm{dx}\:=\:? \\ $$ Commented by i jagooll last updated on 20/May/20 $$\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:.\sqrt{\mathrm{tan}\:\mathrm{x}}}\:=\: \\…
Question Number 160194 by amin96 last updated on 25/Nov/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{6}} }\boldsymbol{\mathrm{dx}}=? \\ $$ Commented by mr W last updated on 25/Nov/21 $$=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 94649 by msup by abdo last updated on 20/May/20 $${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {lnx}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29105 by tawa tawa last updated on 04/Feb/18 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\int_{−\mathrm{1}} ^{\:\:\:\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{dx}}{\mathrm{5}\:\mathrm{cosh}\left(\mathrm{x}\right)\:+\:\mathrm{13}\:\mathrm{sinh}\left(\mathrm{x}\right)}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{log}_{\mathrm{e}} \left(\frac{\mathrm{15e}\:−\:\mathrm{10}}{\mathrm{3e}\:+\:\mathrm{2}}\right)\: \\ $$ Commented by abdo imad last updated on 04/Feb/18 $${I}=\:\int_{−\mathrm{1}}…