Question Number 158288 by cortano last updated on 02/Nov/21 Answered by mr W last updated on 02/Nov/21 $${y}_{\mathrm{1}} −{y}_{\mathrm{2}} ={ax}^{\mathrm{2}} −{ax}−\mathrm{2}{a}={a}\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right) \\ $$$$\Delta={a}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}}…
Question Number 27215 by abdo imad last updated on 03/Jan/18 $${find}\:{the}\:{value}\:{of}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:. \\ $$ Answered by Giannibo last updated on 03/Jan/18 $$…
Question Number 92740 by john santu last updated on 09/May/20 $$\int\:\frac{\mathrm{ln}\left({x}\right)}{\mathrm{cos}\:{x}}\:{dx}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27187 by abdo imad last updated on 02/Jan/18 $${find}\:{I}=\:\:\int_{\mathrm{0}} ^{\propto} \:\frac{{cosx}}{{cosh}\left({x}\right)}{dx} \\ $$ Commented by abdo imad last updated on 08/Jan/18 $${I}=\:\int_{\mathrm{0}} ^{\infty}…
Question Number 27186 by abdo imad last updated on 02/Jan/18 $${find}\:{I}=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{{cosx}\:+\mathrm{2}{sinx}}\:. \\ $$ Commented by abdo imad last updated on 04/Jan/18 $${we}\:{do}\:{the}\:{changement}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\…
Question Number 92723 by Power last updated on 08/May/20 Answered by Rio Michael last updated on 08/May/20 $$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}\:\mathrm{let}\:{x}\:=\:\mathrm{2}\:\mathrm{sec}\:{u} \\ $$$$\Rightarrow\:\frac{{dx}}{{du}}\:=\:\mathrm{2}\:\mathrm{sec}\:{u}\:\mathrm{tan}\:{u} \\ $$$$\Rightarrow\:{dx}\:=\:\mathrm{2}\:\mathrm{sec}\:{u}\:\mathrm{tan}\:{u}\:{du} \\ $$$$\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}}…
Question Number 27185 by abdo imad last updated on 02/Jan/18 $${find}\:\:\int\int_{{D}} \left({x}+{y}\right)^{\mathrm{2}} \:{e}^{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy}\:{with} \\ $$$${D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}\:} /\mathrm{0}<{x}<\mathrm{1}\:{and}\:\mathrm{0}<{y}<\mathrm{1}−{x}\:\right\}. \\ $$ Terms of Service Privacy…
Question Number 27183 by abdo imad last updated on 02/Jan/18 $${find}\:{the}\:{value}\:{of}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}−\mathrm{1}}{{lnt}}{dt}\:. \\ $$ Answered by prakash jain last updated on 03/Jan/18 $${F}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 27184 by abdo imad last updated on 02/Jan/18 $${calculate}\:{in}\:{terms}\:{of}\:{x}\:\:\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \frac{{dt}}{\mathrm{1}+{xsint}}\:. \\ $$ Commented by abdo imad last updated on 05/Jan/18 $${let}\:{do}\:{the}\:{changement}\:\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)=\:\alpha\Leftrightarrow{t}=\mathrm{2}{arctan}\alpha \\…
Question Number 27182 by abdo imad last updated on 02/Jan/18 $$\:{find}\:{the}\:{value}\:{of}\:{I}_{{a}} =\:\int\int_{{D}_{{a}} } {e}^{−\frac{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{\mathrm{2}}} {dxdy}\:\:{with} \\ $$$${D}_{{a}} \:=\left\{\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} \:/\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\:{a}^{\mathrm{2}} \:\:\right\} \\…