Question Number 172944 by flor last updated on 03/Jul/22 $$\int\int_{{d}} {dxdy}\:\:\:\:\:\:{x}={y}^{\mathrm{2}} −\mathrm{1}\:\:\:\:\:\:{x}=\mathrm{1}−{y} \\ $$$${D}=? \\ $$ Commented by kaivan.ahmadi last updated on 04/Jul/22 $${x}={y}^{\mathrm{2}} −\mathrm{1}\Rightarrow{y}^{\mathrm{2}}…
Question Number 41854 by math khazana by abdo last updated on 14/Aug/18 $$\left.\mathrm{1}\right){calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{3}} \right){dx} \\ $$$$\left.\mathrm{2}\right){then}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\:\mathrm{1}−{x}^{\mathrm{5}} \right)\:{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty}…
Question Number 41851 by de best last updated on 13/Aug/18 Commented by math khazana by abdo last updated on 14/Aug/18 $${changement}\:{y}={atan}\left({t}\right)\:{give} \\ $$$$\int\:\:\:\frac{{dy}}{{a}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}}…
Question Number 41848 by maxmathsup by imad last updated on 13/Aug/18 $${let}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\mathrm{1}+{asinx}}\:\:\:{with}\:{a}\in{R} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}+{sinx}}\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+\mathrm{2}{sinx}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…
Question Number 41846 by maxmathsup by imad last updated on 13/Aug/18 $${find}\:\:\int\:\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\:\:+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$ Answered by MJS last updated on 13/Aug/18 $$\int\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{x}^{\mathrm{2}}…
Question Number 41847 by maxmathsup by imad last updated on 13/Aug/18 $${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\frac{{dt}}{{x}\:+{tan}\left({t}\right)} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{anoher}\:{expression}\:{off}\:\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\mathrm{2}+{tan}\left({t}\right)}\:\:\:{and}\:\:{A}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{dt}}{{sin}\theta+{tant}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}}…
Question Number 41845 by maxmathsup by imad last updated on 13/Aug/18 $$\left.\mathrm{1}\right){find}\:\:\:\:\int\:\:\:\:\:\:\:\:\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\:\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$ Answered by MJS last updated on 13/Aug/18…
Question Number 172899 by mathlove last updated on 03/Jul/22 $$\int_{\mathrm{0}} ^{\infty} {e}^{−{e}^{{x}} } \sqrt{{x}}\:{dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107342 by mnjuly1970 last updated on 10/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{E}{valuate}: \\ $$$$\:\:\:\:\:\:\:\chi:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{2}} {tan}\left({x}\right){dx}=\:???\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\bigstar{prepared}\:{by}:\bigstar \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\clubsuit\clubsuit\clubsuit\:\:\:\mathscr{M}.\mathscr{N}\:\clubsuit\clubsuit\clubsuit \\ $$$$ \\ $$ Answered by…
Question Number 41806 by Raj Singh last updated on 13/Aug/18 Commented by prof Abdo imad last updated on 13/Aug/18 $${let}\:{A}\:=\:\int\:\:\frac{{arcsin}\left(.\sqrt{{x}}\right)−{arccos}\left(\sqrt{{x}}\right)}{{arcsin}\sqrt{{x}}\:+{arccos}\sqrt{{x}}}\:{dx} \\ $$$$\left.{arcosx}\:+{arcsinx}\right)^{'} =−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{for}…