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Category: Integration

calculate-I-pi-3-pi-2-cos-2x-sin-x-cosx-dx-and-J-pi-3-pi-2-sin-2x-sin-x-cos-x-dx-

Question Number 42628 by prof Abdo imad last updated on 29/Aug/18 $${calculate}\:\:{I}\:\:=\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{{sin}\left({x}\right)+{cosx}}{dx}\:{and} \\ $$$${J}\:=\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{sin}\left({x}\right)\:+{cos}\left({x}\right)}{dx} \\ $$ Commented by maxmathsup by imad…

find-th-x-dx-

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}}…

let-f-x-arctan-xt-2-1-2t-2-dt-1-find-a-explicite-form-of-f-x-2-calculate-0-arctan-t-2-1-2t-2-dt-and-0-arctan-2t-2-1-2t-2-dt-

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-42593

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-42580

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}}…