Question Number 59659 by aliesam last updated on 13/May/19 Commented by Mr X pcx last updated on 13/May/19 $${yes}\:{sir}. \\ $$ Commented by Mr X…
Question Number 125194 by bemath last updated on 09/Dec/20 $$\:\int\:\frac{\left(\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} \:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:{dx}\:? \\ $$ Answered by liberty last updated on 09/Dec/20 $$\:{by}\:{second}\:{Euler}\:{substitution}\: \\…
Question Number 125187 by bemath last updated on 08/Dec/20 $$\:\:\int\:\frac{{dx}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{3}} {x}}\:\sqrt{\mathrm{cos}\:^{\mathrm{5}} {x}}}\:? \\ $$ Answered by liberty last updated on 08/Dec/20 $${I}=\int\:\sqrt{\frac{\left(\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{3}} {x}}\right)}{\mathrm{tan}\:^{\mathrm{3}} {x}\:\mathrm{cos}\:^{\mathrm{5}} {x}}}\:{dx}\:=\:\int\frac{{dx}}{\:\sqrt{\mathrm{tan}\:^{\mathrm{3}}…
Question Number 59647 by aliesam last updated on 12/May/19 Commented by Mr X pcx last updated on 13/May/19 $${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cosx}\:{e}^{\mid{sinx}\mid} }{\mathrm{1}+{e}^{{tanx}} }\:{dx}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{cosx}\:{e}^{\mid{sinx}\mid}…
Question Number 190708 by mnjuly1970 last updated on 09/Apr/23 Answered by cortano12 last updated on 10/Apr/23 $$\:\mathrm{L}=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:\mathrm{2x}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2x}\right)−\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }} \\ $$$$\:\mathrm{L}=\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{8}.\left(\frac{\mathrm{sin}\:\mathrm{2x}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2x}\right)}{\left(\mathrm{2x}\right)^{\mathrm{3}} }\right)}…
Question Number 59631 by Mr X pcx last updated on 12/May/19 $$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{{acosx}\:+{bsinx}} \\ $$$${with}\:{a}\:,\:{b}\:{reals} \\ $$$$\left.\mathrm{2}\right){find}\:{also}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cosx}\:{dx}}{\left({acosx}\:+{bsinx}\right)^{\mathrm{2}} }\:\:{and} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sinx}\:{dx}}{\left({acosx}\:+{bsinx}\right)^{\mathrm{2}}…
Question Number 190700 by Rupesh123 last updated on 09/Apr/23 Answered by 07049753053 last updated on 09/Apr/23 $$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right){ln}\left({x}\right)}{{x}}{dx} \\ $$$$\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{a}}} }{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{da}}}\mid_{\boldsymbol{\mathrm{a}}=\mathrm{1}}…
Question Number 190692 by mnjuly1970 last updated on 09/Apr/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{sin}^{\:\mathrm{3}} \left({x}\:\right)\:\mathrm{ln}\left(\:{x}\:\right)}{{x}}\:\mathrm{d}{x}\:=\:?\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:@\:\mathrm{nice}\:−\:\mathrm{mathematics}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:…
Question Number 125146 by mathmax by abdo last updated on 08/Dec/20 $$\left.\mathrm{1}\right)\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{d}\theta}{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}\:\mathrm{cos}\theta\:+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{cos}\theta}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\theta\:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{d}\theta \\ $$ Answered by…
Question Number 125133 by mnjuly1970 last updated on 08/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\blacktriangleleft{advanced}\:\:\:{calculus}\blacktriangleright… \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\:::: \\ $$$$\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{{cos}\left({log}\left({x}\right)\right)−\mathrm{1}}{{log}\left({x}\right)}\right\}{dx}=\frac{{log}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:…\ast{adopted}\:{from}\:{youtube}\ast…\:\:\: \\ $$$$\:\ast\:\ast\:{youtube}\:{solution}\:{is}\:{not}\:{considered}\:\ast\:\ast \\ $$$$\:\: \\ $$ Answered…