Question Number 40829 by math khazana by abdo last updated on 28/Jul/18 $${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({tx}\right)}{{x}^{\mathrm{3}} +\mathrm{8}}{dx} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}\right)}{{x}^{\mathrm{3}} \:+\mathrm{8}}{dx}\:. \\ $$…
Question Number 106365 by Mikael_786 last updated on 04/Aug/20 $$\mathrm{help} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)!} \\ $$ Commented by Dwaipayan Shikari last updated on 04/Aug/20 $${or}\:{sin}\:{h}\left(\mathrm{1}\right)…
Question Number 40823 by math khazana by abdo last updated on 28/Jul/18 $${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}^{\mathrm{2}} {x}\:+\mathrm{3}{sin}^{\mathrm{2}} {x}}{dx} \\ $$ Commented by maxmathsup by imad last…
Question Number 40787 by rahul 19 last updated on 27/Jul/18 $$\mathrm{Let}\:\mathrm{I}_{\mathrm{1}} =\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{sin}\:{x}}{{x}}\:{dx}\:\:,\:\:\mathrm{I}_{\mathrm{2}} =\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{sin}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:{x}}{dx} \\ $$$$,\:\mathrm{I}_{\mathrm{3}} =\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)}{\mathrm{tan}\:{x}}{dx}.\: \\ $$$${P}\mathrm{rove}\:\mathrm{that}\:\mathrm{I}_{\mathrm{2}} \:>\:\mathrm{I}_{\mathrm{1}}…
Question Number 40760 by math khazana by abdo last updated on 27/Jul/18 $${find}\:\:\int\:\:\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 106285 by bemath last updated on 04/Aug/20 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:? \\ $$ Answered by bobhans last updated on 04/Aug/20 $$\mathrm{recall}\:\underset{−\infty} {\overset{\infty}…
Question Number 40745 by Raj Singh last updated on 27/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18 $${x}^{\mathrm{3}} ={a}^{\mathrm{3}} {sin}^{\mathrm{2}} \alpha \\ $$$$\mathrm{3}{x}^{\mathrm{2}} {dx}={a}^{\mathrm{3}} .\mathrm{2}{sin}\alpha{cos}\alpha\:{d}\alpha…
Question Number 40716 by ajeetyadav4370 last updated on 26/Jul/18 $$\int\left({cosx}−{cos}\mathrm{2}{x}/\mathrm{1}−{cosx}\right){dx} \\ $$ Answered by maxmathsup by imad last updated on 26/Jul/18 $${let}\:{I}\:=\:\int\:\:\frac{{cosx}\:−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{cosx}}\:{dx} \\ $$$${I}\:=\:\int\:\:\:\frac{{cosx}\:−\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)}{\mathrm{1}−{cosx}}{dx}\:=\:\int\:\:\frac{−\mathrm{2}{cos}^{\mathrm{2}}…
Question Number 40717 by ajeetyadav4370 last updated on 26/Jul/18 $$\int\sqrt{{tanx}/{sinx}.{cosxdx}} \\ $$ Commented by math khazana by abdo last updated on 26/Jul/18 $${let}\:{I}\:\:=\:\int\:\:\sqrt{\frac{{tanx}}{{sinx}\:{cosx}}}{dx} \\ $$$${I}\:=\:\int\:\:\:\sqrt{\frac{{sinx}}{{sinx}\:{cos}^{\mathrm{2}}…
Question Number 106232 by abdomathmax last updated on 03/Aug/20 $$\mathrm{find}\:\int\:\mathrm{cos}^{\mathrm{n}} \mathrm{x}\:\mathrm{ch}\left(\mathrm{nx}\right)\mathrm{dx}\:/\mathrm{with}\:\mathrm{n}\:\mathrm{integr} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com