Question Number 54102 by gunawan last updated on 29/Jan/19 $$\mathrm{the}\:\mathrm{absolute}\:\mathrm{value} \\ $$$$\int_{\mathrm{10}} ^{\mathrm{19}} \frac{\mathrm{cos}\:{x}}{\mathrm{1}+{x}^{\mathrm{8}} }\:{dx}\:\mathrm{is}… \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19 $${from}\:{enclosed}\:{graph}\:{below}\:{it}\:{is}\:{clear}\:{that}…
Question Number 54074 by rahul 19 last updated on 28/Jan/19 $${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right)\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}+\mathrm{2}}\: \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right) \\…
Question Number 54070 by rahul 19 last updated on 28/Jan/19 $${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right) \\ $$$$\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right).\left(\mathrm{co}{t}^{−\mathrm{1}} \frac{{x}}{\:\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mid{x}\mid} }}\right){dx} \\ $$$$\left.\mathrm{2}\right) \\…
Question Number 54073 by cesar.marval.larez@gmail.com last updated on 28/Jan/19 Answered by estudiante last updated on 28/Jan/19 $${Vemos}\:{q}\:{es}\:{una}\:{integral}\:{impropia}\:{de}\:{tipo}\:{I}: \\ $$$$\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\int_{{a}} ^{{R}} {x}^{{n}} {dx}\:=\:\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\mid\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mid_{{a}}…
Question Number 119591 by Lordose last updated on 25/Oct/20 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$ \\ $$$$\:\: \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\left(\mathrm{2}^{{x}} −\mathrm{1}\right){sin}^{\mathrm{3}} \left({x}\right)}{\left(\mathrm{2}^{{x}} +\mathrm{1}\right)\left({sin}^{\mathrm{3}}…
Question Number 119570 by mnjuly1970 last updated on 25/Oct/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}{nice}\:\:{calculus}\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}… \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {e}^{−\mathrm{2}{x}} {ln}\left(\frac{\mathrm{1}+{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\bigstar\:\mathscr{M}.\mathscr{N}.\mathrm{1970}\bigstar……
Question Number 185104 by emmanuelson123 last updated on 17/Jan/23 Answered by aba last updated on 17/Jan/23 $$\pi\mathrm{ln}\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right) \\ $$ Commented by emmanuelson123 last updated on…
Question Number 54011 by maxmathsup by imad last updated on 27/Jan/19 $${calculate}\:{f}\left({a}\right)\:=\int\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{ax}}−\sqrt{\mathrm{1}−{ax}}}\:\:{with}\:{a}>\mathrm{0}\:. \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:{U}_{{n}} =\int_{−\frac{\mathrm{1}}{{na}}} ^{\frac{\mathrm{1}}{{na}}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{ax}}−\sqrt{\mathrm{1}−{ax}}}\:\:{with}\:{n}\:{from}\:{N}\:{and}\:{n}>\mathrm{1} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \:\:\:{and}\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$ Commented by…
Question Number 185058 by saboorhalimi last updated on 16/Jan/23 Answered by Frix last updated on 16/Jan/23 $${k}\mathrm{cos}\:{nx}\:+{m}\mathrm{sin}\:{nx}\:=\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} }\mathrm{sin}\:\left({nx}+\mathrm{tan}^{−\mathrm{1}} \:\frac{{k}}{{m}}\right) \\ $$$$\Omega=\sqrt{{k}^{\mathrm{2}} +{m}^{\mathrm{2}} }\underset{\mathrm{0}} {\overset{\pi}…
Question Number 53967 by maxmathsup by imad last updated on 27/Jan/19 $$\left.\mathrm{1}\right){calculate}\:{A}_{{t}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{sinxdx}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{by}\:{using}\:{Fubuni}\:{theorem}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}}{dx}\:. \\ $$ Commented by maxmathsup…