Question Number 42622 by maxmathsup by imad last updated on 29/Aug/18 $${find}\:\int\:\:{th}\left(\mathrm{2}{x}+\mathrm{1}\right){dx}\: \\ $$ Commented by maxmathsup by imad last updated on 31/Aug/18 $${changement}\:\mathrm{2}{x}+\mathrm{1}\:={t}\:{give} \\…
Question Number 42621 by maxmathsup by imad last updated on 29/Aug/18 $${find}\:\int\:{th}\left({x}\right){dx}\: \\ $$ Commented by maxmathsup by imad last updated on 30/Aug/18 $${let}\:{I}\:\:=\:\int\:{th}\left({x}\right)\Rightarrow\:{I}\:=\int\:\:\:\frac{{sh}\left({x}\right)}{{ch}\left({x}\right)}{dx}\:=\int\:\:\:\frac{{e}^{{x}} −{e}^{−{x}}…
Question Number 108144 by mohammad17 last updated on 15/Aug/20 $$\int\:\sqrt{{ln}\left({x}\right)}{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42605 by maxmathsup by imad last updated on 28/Aug/18 $${let}\:{f}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{arctan}\:\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicite}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{arctan}\left({t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{dt}\:\:{and}\:\int_{\mathrm{0}} ^{\infty}…
Question Number 42603 by maxmathsup by imad last updated on 28/Aug/18 $${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{ixt}\right){dt}\:\:{calculate}\:{f}^{,} \left({x}\right)\:\:\:\:\left({x}\:{from}\:{R}\right). \\ $$ Commented by prof Abdo imad last updated on…
Question Number 42593 by aseerimad last updated on 28/Aug/18 Commented by maxmathsup by imad last updated on 28/Aug/18 $${we}\:{have}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{cos}^{\mathrm{4}} {x}\:+{sin}^{\mathrm{4}} \:{x}−{sin}^{\mathrm{4}} {x}}{{sin}^{\mathrm{4}} {x}\:+{cos}^{\mathrm{4}}…
Question Number 42580 by Raj Singh last updated on 28/Aug/18 Commented by maxmathsup by imad last updated on 28/Aug/18 $${changement}\:\sqrt{{x}+\mathrm{1}}={t}\:{give}\:{x}={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow \\ $$$$\int\:\:\:\frac{{x}+\sqrt{{x}+\mathrm{1}}}{{x}+\mathrm{2}}{dx}\:=\:\int\:\:\:\frac{{t}^{\mathrm{2}} −\mathrm{1}\:+{t}}{{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{2}}\left(\mathrm{2}{t}\right){dt}\:=\:\int\:\:\:\:\frac{\mathrm{2}{t}^{\mathrm{3}}…
Question Number 42579 by Raj Singh last updated on 28/Aug/18 $$\int\:{sinx}/\sqrt{\mathrm{1}+{sinx}} \\ $$ Commented by maxmathsup by imad last updated on 28/Aug/18 $${let}\:{A}\:=\:\int\:\:\:\frac{{sinx}}{\:\sqrt{\mathrm{1}+{sinx}}}\:{dx}\:\:{we}\:{have}\:\:\mathrm{1}+{sinx}\:=\mathrm{1}+\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}}…
Question Number 108099 by bemath last updated on 14/Aug/20 $$\:\:\:\:\frac{\bigstar\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{\sqcap} \\ $$$$\:\left(\mathrm{1}\right)\:\int\:{x}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\:{dx}\:? \\ $$$$\left(\mathrm{2}\right)\:{Find}\:{the}\:{distance}\:{of}\:{the}\:{point}\: \\ $$$$\left(\mathrm{3},\mathrm{3},\mathrm{1}\right)\:{from}\:{the}\:{plane}\:\Pi\:{with}\:{equation} \\ $$$$\left(\overset{\rightarrow} {{r}}−\overset{\rightarrow} {{i}}−\overset{\rightarrow} {{j}}\right)\bullet\left(\overset{\rightarrow} {{i}}−\overset{\rightarrow} {{j}}+\overset{\rightarrow} {{k}}\right)\:=\:\mathrm{0}\:,\:{also}\:…
Question Number 108095 by bemath last updated on 14/Aug/20 $$\:\:\:\frac{\infty\mathcal{B}{e}\mathcal{M}{ath}\infty}{\spadesuit} \\ $$$$\:\:\:\int\:\mathrm{2}{x}\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\:{dx}\: \\ $$ Answered by bemath last updated on 14/Aug/20 Answered by Dwaipayan…