Question Number 179433 by Spillover last updated on 29/Oct/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 179405 by Acem last updated on 29/Oct/22 $$\:{Evaluate}\:\int\:\frac{{dx}}{{x}^{\mathrm{4}} \:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$ Answered by greougoury555 last updated on 29/Oct/22 $$\:{M}=\int\:\frac{{dx}}{{x}^{\mathrm{3}} \:{x}^{\mathrm{2}} \:\sqrt{\frac{\mathrm{9}}{{x}^{\mathrm{2}} }−\mathrm{1}}}\:{dx}…
Question Number 113867 by Ar Brandon last updated on 15/Sep/20 $$\int_{\mathrm{3}} ^{\mathrm{6}} \frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{6x}}\mathrm{dx} \\ $$ Answered by abdomsup last updated on 15/Sep/20 $${I}\:=\int_{\mathrm{3}}…
Question Number 113868 by Ar Brandon last updated on 15/Sep/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:\mathrm{circle}\:\rho=\mathrm{2acos}\theta\:\mathrm{and}\: \\ $$$$\mathrm{cardiode}\:\rho=\mathrm{a}\left(\mathrm{1}+\mathrm{cos}\theta\right) \\ $$ Commented by kaivan.ahmadi last updated on 18/Sep/20 $$\mathrm{2}{acos}\theta={a}+{acos}\theta\Rightarrow{acos}\theta={a}\Rightarrow{cos}\theta=\mathrm{1}\Rightarrow \\ $$$$\theta=\mathrm{0},\theta=\mathrm{2}\pi…
Question Number 113865 by Ar Brandon last updated on 15/Sep/20 $$\mathrm{Consider}\:\mathrm{the}\:\mathrm{series}\:\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{1}} ^{\mathrm{e}} \mathrm{x}\left(\mathrm{lnx}\right)^{\mathrm{n}} \mathrm{dx}\:\mathrm{and}\:\mathrm{I}_{\mathrm{0}} =\int_{\mathrm{1}} ^{\mathrm{e}} \mathrm{xdx} \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true}\:? \\ $$$$\mathrm{a}\backslash\:\mathrm{0}\leqslant\mathrm{I}_{\mathrm{n}} \leqslant\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\:\:\:\:\mathrm{b}\backslash\mathrm{1}\leqslant\mathrm{I}_{\mathrm{n}} \leqslant\frac{\mathrm{e}^{\mathrm{2}}…
Question Number 179383 by Acem last updated on 29/Oct/22 $$\mathrm{1}\bullet\:{Evaluate}\:{I}=\int\:\:\frac{\sqrt{\mathrm{25}{x}^{\mathrm{2}} −\mathrm{4}}}{{x}}\:{dx} \\ $$$$\:\mathrm{2}\bullet\:{Find}\:{value}\:{of}\:\int_{\frac{\mathrm{2}}{\mathrm{5}}} ^{\:\frac{\mathrm{4}}{\mathrm{5}}} {f}\left({x}\right)\:{dx} \\ $$$$\:\mathrm{3}\bullet\:{Find}\:{value}\:{of}\:\int_{−\:\frac{\mathrm{4}}{\mathrm{5}}} ^{\:−\:\frac{\mathrm{2}}{\mathrm{5}}} {f}\left({x}\right)\:{dx} \\ $$$$ \\ $$ Answered by…
Question Number 48289 by Abdulhafeez Abu qatada last updated on 21/Nov/18 $${Evaluate}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{Log}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\:{dx} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 22/Nov/18 Commented…
Question Number 113821 by 675480065 last updated on 15/Sep/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{2}−\mathrm{sinx}\right)\mathrm{dx} \\ $$ Commented by Dwaipayan Shikari last updated on 15/Sep/20 $${I}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\mathrm{2}+{asinx}\right){dx}…
Question Number 179344 by neinhaltsieger369 last updated on 28/Oct/22 $$\: \\ $$$$\:\mathrm{Help}-\mathrm{me}! \\ $$$$\: \\ $$$$\:\int_{\mathrm{0}} ^{\:\boldsymbol{\pi}} \int_{\mathrm{0}} ^{\:\mathrm{3}\boldsymbol{\mathrm{cos}}\:\boldsymbol{\phi}} \boldsymbol{\theta\mathrm{sin}}\:\boldsymbol{\phi\mathrm{d}\theta\mathrm{d}\phi} \\ $$$$\: \\ $$ Commented…
Question Number 48264 by Abdo msup. last updated on 21/Nov/18 $${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{x}\:{cos}\left({t}\right)}{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{also}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{sin}\left(\mathrm{2}{t}\right){cost}}{\left(\mathrm{1}+{xcost}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{\mathrm{1}+\mathrm{3}\:{cos}\left({t}\right)}{dt}\:{and} \\…