Question Number 67937 by A8;15: last updated on 02/Sep/19 Commented by mathmax by abdo last updated on 02/Sep/19 $${at}\:{form}\:{of}\:{serie} \\ $$$${I}\:=\int\sqrt{{e}^{{x}} }{dx}\:=\:\int\:\:{e}^{\frac{{x}}{\mathrm{2}}} {dx}\:=\int\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(\frac{{x}}{\mathrm{2}}\right)^{{n}}…
Question Number 133470 by rs4089 last updated on 22/Feb/21 Answered by mr W last updated on 22/Feb/21 $${x}={a}\:\mathrm{cos}\:\theta \\ $$$${y}={b}\:\mathrm{sin}\:\theta \\ $$$${dV}=−\mathrm{2}\pi\left(\mathrm{2}{a}−{a}\:\mathrm{cos}\:\theta\right)\mathrm{2}{ydx} \\ $$$$=\mathrm{4}\pi{a}\left(\mathrm{2}−\mathrm{cos}\:\theta\right){b}\mathrm{sin}\:\theta{a}\mathrm{sin}\:\theta{d}\theta \\…
Question Number 67932 by mathmax by abdo last updated on 02/Sep/19 $${let}\:{A}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}^{\mathrm{4}} −{e}^{{i}\theta} \right)}\:\:{with}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}\left(\theta\right)\:{interms}\:{of}\:\theta \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{also}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}^{\mathrm{4}} −{e}^{{i}\theta}…
Question Number 67931 by mathmax by abdo last updated on 02/Sep/19 $${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{x}^{{n}} \left\{\mathrm{1}+{cosx}\:+{cos}\left(\mathrm{2}{x}\right)\right\}^{\mathrm{2}} {dx} \\ $$$${find}\:{a}\:{relation}\:{of}\:{recurrence}\:{betwedn}\:{the}\:{A}_{{n}} \\ $$ Terms of Service Privacy…
Question Number 67921 by mhmd last updated on 02/Sep/19 $$\int{e}^{{y}^{\mathrm{2}} /\mathrm{2}} {dy} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133452 by bagjagunawan last updated on 22/Feb/21 Answered by mnjuly1970 last updated on 22/Feb/21 $$\boldsymbol{\phi}\overset{{x}=\mathrm{2}{y}−\mathrm{1}} {=}\int_{\frac{\mathrm{1}}{\mathrm{2}}\:} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{2}−\mathrm{2}{y}\right){ln}\left({ln}\left(\mathrm{2}{y}\right)\right)}{\mathrm{2}{y}}\left(\mathrm{2}\right){dy} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \left({ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−{y}\right)\right)\left({ln}\left(\mathrm{2}\right)+{ln}\left({y}\right)\right)\frac{{dy}}{{y}} \\…
Question Number 67919 by aliesam last updated on 02/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 67907 by A8;15: last updated on 02/Sep/19 Commented by mathmax by abdo last updated on 02/Sep/19 $${let}\:{I}\:=\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{6}}} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0}\rightarrow\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}\right)\:=\mathrm{25}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{5}}{\mathrm{2}}=\mathrm{2}\:\:{and}\: \\…
Question Number 133440 by bagjagunawan last updated on 22/Feb/21 Answered by Ar Brandon last updated on 22/Feb/21 $$\mathrm{f}\left({x}\right)={x}+\mathrm{5}+\sqrt{\mathrm{8}{x}}+\sqrt{\mathrm{12}{x}}+\sqrt{\mathrm{24}}\: \\ $$$$\:\:\:\:\:\:\:\:\:={x}+\left(\sqrt{\mathrm{8}}+\sqrt{\mathrm{12}}\right)\sqrt{{x}}+\mathrm{5}+\sqrt{\mathrm{24}}=\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\mathcal{I}=\int\mathrm{f}\left({x}\right)\mathrm{d}{x}=\int\sqrt{\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\mathrm{d}{x} \\ $$$$\:\:\:=\int\left(\sqrt{{x}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\mathrm{d}{x}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{x}^{\mathrm{3}}…
Question Number 133433 by greg_ed last updated on 22/Feb/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}, \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\::\:\mathrm{2}\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{1}}\:\boldsymbol{{dx}}. \\ $$ Answered by…