Question Number 27802 by abdo imad last updated on 15/Jan/18 $${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{x}^{\mathrm{2}} } }{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:. \\ $$ Commented by abdo imad last updated…
Question Number 27803 by abdo imad last updated on 15/Jan/18 $${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({x}\:+{x}^{−\mathrm{1}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27797 by abdo imad last updated on 14/Jan/18 $${find}\:\:\:\int\:\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {t}}\:\:{dt}. \\ $$ Commented by abdo imad last updated on 19/Jan/18 $${let}\:{put}\:\:{I}=\:\int\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} {t}}{dt}\:\:\:\: \\…
Question Number 27796 by abdo imad last updated on 14/Jan/18 $${find}\:\:\int\:\:\:\frac{{x}^{\mathrm{2}} }{\left({cosx}\:+{x}\:{sinx}\right)^{\mathrm{2}} }\:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27794 by abdo imad last updated on 14/Jan/18 $${let}\:{give}\:\:{I}\left({x}\right)=\:\int_{\mathrm{0}} ^{\pi} {ln}\:\left(\mathrm{1}−\mathrm{2}{x}\:{cost}\:+{x}^{\mathrm{2}} \right){dt}\:{by}\:{using}\:{the} \\ $$$${polynomial}\:{p}\left({x}\right)=\:\left({z}+{x}\right)^{\mathrm{2}{n}} −\mathrm{1}\:\:{find}\:{the}\:{value}\:{of}\:{I}\left({x}\right). \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 27788 by abdo imad last updated on 14/Jan/18 $${find}\:{the}\:{value}\:{of}\:\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sin}\left({nt}\right)}{{sint}}{dt}\:\:{with}\:{n}\in{N}^{\ast} \:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27781 by abdo imad last updated on 14/Jan/18 $${find}\:{the}\:{value}\:{of}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{ln}\left(\mathrm{1}+{x}\:{sin}^{\mathrm{2}} {t}\right)}{{sin}^{\mathrm{2}} {t}}\:{dt}\:{knowing}\:{that} \\ $$$$−\mathrm{1}<{x}<\mathrm{1}\:. \\ $$ Commented by abdo imad last updated…
Question Number 27764 by NECx last updated on 14/Jan/18 $$\int\sqrt{\mathrm{tan}\:{x}}{dx} \\ $$ Answered by prakash jain last updated on 14/Jan/18 Terms of Service Privacy Policy…
Question Number 158839 by EbrimaDanjo last updated on 09/Nov/21 $$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integrals}\:\mathrm{using} \\ $$$$\mathrm{integration}\:\boldsymbol{\mathrm{By}}\:\boldsymbol{\mathrm{Parts}} \\ $$$$\mathrm{1}.\:\int_{\frac{\pi\:}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {xcsc}^{\mathrm{2}} {xdx} \\ $$$$ \\ $$$$\mathrm{2}.\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{3}}} {arctan}\left(\frac{\mathrm{1}}{{x}}\right){dx} \\ $$…
Question Number 27757 by abdo imad last updated on 14/Jan/18 $${calculate}\:\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+{cosx}}\:{and}\:{J}=\:\int_{\mathrm{0}^{} } ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cosx}}{\mathrm{1}+{cosx}}{dx}\:. \\ $$ Answered by mrW2 last updated on 14/Jan/18…