Question Number 151851 by Tawa11 last updated on 23/Aug/21 Answered by OlafThorendsen last updated on 23/Aug/21 $$\mathrm{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}} \left({x}^{\mathrm{2}} +\mathrm{ln}\left(\frac{\pi+{x}}{\pi−{x}}\right)\right)\mathrm{cos}{x}\:{dx}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{u}\:=\:−{x}\:: \\ $$$$\mathrm{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}}…
Question Number 86313 by john santu last updated on 28/Mar/20 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}} \\ $$ Commented by john santu last updated on 28/Mar/20 $$\frac{\sqrt{\mathrm{sin}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:+\:\frac{\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:=\:\mathrm{1} \\…
Question Number 151838 by talminator2856791 last updated on 23/Aug/21 $$\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\left({x}^{\mathrm{log}\left(\lfloor\left(\lfloor{x}\rfloor!\right)^{\left(\mathrm{log}\left(\lfloor{x}−\mathrm{1}\rfloor!\right)\right)^{−\mathrm{1}} } \rfloor\right)+\mathrm{1}} +\mathrm{1}\right)^{{x}} }{\lfloor{x}^{\mathrm{log}\left({x}^{{x}} \right)+\mathrm{1}} \rfloor!+\mathrm{1}}\:{dx} \\ $$$$\: \\ $$ Terms…
Question Number 86302 by john santu last updated on 28/Mar/20 $$\int\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}\:=? \\ $$ Commented by abdomathmax last updated on 28/Mar/20 $${I}\:=\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}\:{changement}\:{x}\:=\mathrm{2}{sh}\left({t}\right)\:{give}…
Question Number 20733 by Joel577 last updated on 02/Sep/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{of}\:\mathrm{revolution} \\ $$$$\mathrm{obtained}\:\mathrm{by}\:\mathrm{revolving}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by} \\ $$$${x}\:=\:\mathrm{4}\:+\:\mathrm{6}{y}\:−\:\mathrm{2}{y}^{\mathrm{2}} ,\:{x}\:=\:−\mathrm{4},\:{x}\:=\:\mathrm{0}\:\mathrm{about} \\ $$$$\mathrm{the}\:{y}−\mathrm{axis} \\ $$ Answered by mrW1 last updated on…
Question Number 86269 by M±th+et£s last updated on 27/Mar/20 $$\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\left({tan}\left({x}\right)−\mathrm{1}\right)^{\mathrm{4}} \left({tan}\left({x}\right)−\mathrm{2}\right)}\:{dx} \\ $$ Commented by john santu last updated on 28/Mar/20 $${u}\:=\:\mathrm{tan}\:{x}−\mathrm{2}\: \\ $$$$\Rightarrow\:\int\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{4}}…
Question Number 86258 by lémùst last updated on 27/Mar/20 $${calculate}\:{I}=\int_{\mathrm{1}} ^{+\infty} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 86254 by sakeefhasan05@gmail.com last updated on 27/Mar/20 $$\int\:\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}…….}}}}\:\:\:\mathrm{dx} \\ $$ Answered by TANMAY PANACEA. last updated on 27/Mar/20 $${y}=\sqrt{{x}\sqrt{{x}\sqrt{{x}\sqrt{{x}…\infty}\:\:\:}}} \\ $$$${y}=\sqrt{{xy}}\: \\ $$$${y}^{\mathrm{2}}…
Question Number 86246 by Rio Michael last updated on 27/Mar/20 $$\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:{dx}\:=\:? \\ $$ Answered by TANMAY PANACEA. last updated on 27/Mar/20 $$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 86247 by M±th+et£s last updated on 27/Mar/20 $$\int_{\mathrm{0}} ^{\mathrm{8}} \int_{\mathrm{0}} ^{{x}^{\frac{\mathrm{2}}{\mathrm{3}}} } \:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}}\:{dy}\:{dx} \\ $$ Commented by mathmax by abdo last…