Question Number 108663 by bobhans last updated on 18/Aug/20 Answered by john santu last updated on 18/Aug/20 $$\:\:\:\:\:\frac{\frac{\multimap{J}}{{S}\leftrightharpoons}}{} \\ $$$${let}\:\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }\:=\:{m}\:\Rightarrow{x}^{\mathrm{3}} \:=\:{m}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}}…
Question Number 43125 by Raj Singh last updated on 07/Sep/18 Commented by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{I}\:=\int\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{{sinx}}+\frac{\mathrm{1}}{{cosx}}}{dx}\:\Rightarrow{I}\:=\:\int\:\:\:\:\frac{{cosx}\:{sinx}}{{cosx}\:+{sinx}}{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{2}}{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{dx}\:\:{changement}\:\:{x}−\frac{\pi}{\mathrm{4}}={t}\:{give}\: \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\frac{{sin}\left(\mathrm{2}\left({t}+\frac{\pi}{\mathrm{4}}\right)\right)}{{cost}}\:{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{{cost}}\:{dt} \\…
Question Number 43100 by maxmathsup by imad last updated on 07/Sep/18 $${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cos}\theta}{\mathrm{1}+{xsin}\theta}{d}\theta \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}\left(\mathrm{2}\theta\right)}{\left(\mathrm{1}+{xsin}\theta\right)^{\mathrm{2}} }{d}\theta \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{cos}\theta}{\mathrm{1}+\mathrm{2}{cos}\theta}{d}\theta\:\:\:{and}\:\:\int_{\mathrm{0}}…
Question Number 108605 by bemath last updated on 18/Aug/20 $$\:\:\:\frac{\boldsymbol{{B}}{e}\boldsymbol{{M}}{ath}}{\approxeq} \\ $$$$\:\int\:\frac{{x}^{\mathrm{11}} }{\left({x}^{\mathrm{8}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\: \\ $$ Answered by bemath last updated on 18/Aug/20 $$\:\:{let}\:{x}^{\mathrm{4}}…
Question Number 174133 by mnjuly1970 last updated on 26/Jul/22 Commented by MJS_new last updated on 25/Jul/22 $$\mathrm{but}\:\Omega=\pi \\ $$ Commented by mnjuly1970 last updated on…
Question Number 43064 by abdo.msup.com last updated on 07/Sep/18 Commented by behi83417@gmail.com last updated on 07/Sep/18 $${hi}\:{mr}.{prop}.\:{abdo}.\:{glad}\:{to}\:{see}\:{you}. \\ $$$${good}\:{luck}. \\ $$ Commented by abdo.msup.com last…
Question Number 108597 by bemath last updated on 18/Aug/20 $$\:\:\:\:\:\frac{\angle\:\mathcal{B}{e}\mathcal{M}{ath}\angle}{\nparallel} \\ $$$$\left(\mathrm{1}\right)\:\int\:\mathrm{cos}\:\left(\mathrm{ln}\:{x}\right)\:{dx}\: \\ $$$$\left(\mathrm{2}\right)\:\int\:\mathrm{sin}\:\left(\mathrm{ln}\:{x}\right)\:{dx}\: \\ $$ Answered by 1549442205PVT last updated on 18/Aug/20 $$\mathrm{Set}\:\mathrm{I}=\int\mathrm{cos}\left(\mathrm{lnx}\right)\mathrm{dx},\mathrm{J}=\int\mathrm{sin}\left(\mathrm{lnx}\right)\mathrm{dx} \\…
Question Number 43058 by maxmathsup by imad last updated on 06/Sep/18 $${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{x}\:{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\left\{\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right\}\left\{\mathrm{1}+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right\}}{dx} \\ $$ Commented by maxmathsup by imad last updated…
Question Number 43057 by maxmathsup by imad last updated on 06/Sep/18 $${find}\:{the}\:{value}\:{of}\:\:\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\pi{x}^{\mathrm{2}} \right)\:−{sin}\left(\pi{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$ Commented by maxmathsup by imad…
Question Number 108593 by Rasikh last updated on 18/Aug/20 Answered by Her_Majesty last updated on 18/Aug/20 $$\left(−\mathrm{1}\right)^{{x}} ={e}^{{xln}\left(−\mathrm{1}\right)} ={e}^{{i}\pi{x}} ={cos}\pi{x}+{isin}\pi{x} \\ $$$$\Rightarrow\:\int\left(−\mathrm{1}\right)^{{x}} {dx}=\frac{\mathrm{1}}{\pi}\left({sin}\pi{x}−{icos}\pi{x}\right)+{C} \\ $$…