Question Number 173859 by mathlove last updated on 20/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42773 by maxmathsup by imad last updated on 02/Sep/18 $${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{e}^{{t}} }{\mathrm{1}+{x}^{{t}} }\:{dt}\:\:\:\:\:{with}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${give}\:{f}\left({x}\right)\:{at}\:{form}\:{of}\:{serie}\:. \\ $$ Terms of Service Privacy Policy…
Question Number 42771 by maxmathsup by imad last updated on 02/Sep/18 $$\left.\mathrm{1}\right)\:{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$ Terms of Service…
Question Number 108306 by Sayantan chakraborty last updated on 16/Aug/20 $$\mathrm{proof}\:\mathrm{that}\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}{\left({sin}^{\mathrm{3}} {x}+{cos}^{\mathrm{3}} {x}\right)^{\mathrm{2}} }\:{dx}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Terms of Service Privacy Policy…
Question Number 42770 by maxmathsup by imad last updated on 02/Sep/18 $$\left.\mathrm{1}\right){find}\:{A}\left(\xi\right)\:=\:\int_{\mathrm{0}} ^{\xi} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right){dx}\:\:{with}\:\:\mathrm{0}<\xi<\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right){dx} \\ $$ Terms of Service Privacy Policy…
Question Number 42769 by maxmathsup by imad last updated on 02/Sep/18 $${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{xe}^{−{x}} }\:{dx}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 108305 by Sayantan chakraborty last updated on 16/Aug/20 $${proof}\:{that}\underset{−\infty} {\overset{\infty} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} +{ax}+{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{bx}+{b}^{\mathrm{2}} \right)}\:{equals}\:\:\frac{\mathrm{2}\pi\left(\mathrm{a}+\mathrm{b}\right)}{\:\sqrt{}\mathrm{3ab}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \right)}. \\ $$ Terms of Service Privacy…
Question Number 108303 by Sayantan chakraborty last updated on 16/Aug/20 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{\mathrm{1}}{\left({sinx}+{cosx}\right)}{sin}^{\mathrm{6}} {x}\:{dx}\:{equals} \\ $$ Answered by Sayantan chakraborty last updated on 16/Aug/20 $$\mathrm{Answer}:\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\left(\sqrt{}\mathrm{2}+\mathrm{1}\right)+\mathrm{1}\right).…
Question Number 108285 by Ar Brandon last updated on 16/Aug/20 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{t}^{\mathrm{5}} +\mathrm{3t}+\mathrm{1}}{\mathrm{t}^{\mathrm{3}} +\mathrm{100}}\mathrm{e}^{−\mathrm{t}} \mathrm{dt} \\ $$ Terms of Service Privacy Policy…
Question Number 108283 by Ar Brandon last updated on 16/Aug/20 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:\Gamma\:\mathrm{defined}\:\mathrm{by}\:\Gamma\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{+\infty} \mathrm{t}^{\mathrm{x}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$\mathrm{1}.\:\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{definition}\:\mathrm{of}\:\Gamma\:? \\ $$$$\mathrm{2}.\:\:\mathrm{Show}\:\mathrm{that}\:\forall\mathrm{x}\in\:\mathrm{D}\Gamma,\:\mathrm{x}\Gamma\left(\mathrm{x}\right)=\Gamma\left(\mathrm{x}+\mathrm{1}\right)\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\Gamma\left(\mathrm{n}\right),\:\mathrm{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{3}.\:\:\mathrm{Assuming}\:\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } =\frac{\sqrt{\pi}}{\mathrm{2}},\:\mathrm{calculate}\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{that}…