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Category: Integration

Question-103863

Question Number 103863 by mohammad17 last updated on 17/Jul/20 Answered by Dwaipayan Shikari last updated on 17/Jul/20 $$\left.\mathrm{2}\right) \\ $$$$\int\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{dx}=\int\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{−\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$={sin}^{−\mathrm{1}}…

dx-x-1-3-2-

Question Number 103846 by Dwaipayan Shikari last updated on 17/Jul/20 $$\int\frac{{dx}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2}} \\ $$ Commented by Dwaipayan Shikari last updated on 17/Jul/20 $$\int\frac{\mathrm{3}{u}^{\mathrm{2}} {du}}{{u}+\mathrm{2}}=\int\frac{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{12}}{{u}+\mathrm{2}}+\frac{\mathrm{12}}{{u}+\mathrm{2}}=\int\mathrm{3}{u}−\mathrm{6}+\mathrm{12}{log}\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}}…

let-f-x-0-arctan-xt-1-t-2-dt-with-x-0-1-calculate-f-x-then-a-simple-form-of-f-x-2-calculate-0-arctant-1-t-2-dt-3-calculate-0-arctan-2t-1-t-

Question Number 38310 by maxmathsup by imad last updated on 23/Jun/18 $${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctan}\left({xt}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{then}\:{a}\:{simple}\:{form}\:{of}\:\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctant}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}}…

dx-1-sinx-2-

Question Number 103825 by mohammad17 last updated on 17/Jul/20 $$\int\:\frac{{dx}}{\left(\mathrm{1}−{sinx}\right)^{\mathrm{2}} }\:? \\ $$ Answered by ~blr237~ last updated on 17/Jul/20 $$\:\int\:\frac{\mathrm{1}+\mathrm{2}{sinx}+\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)}{{cos}^{\mathrm{4}} {x}}{dx} \\ $$$$\int\:\left[\:\mathrm{2}\left(\mathrm{1}+{tan}^{\mathrm{2}}…

c-x-2-2xy-2-dx-x-2-y-2-1-dy-where-C-is-the-boundary-of-region-define-by-y-2-4x-and-y-1-

Question Number 103773 by bemath last updated on 17/Jul/20 $$\int_{{c}} \left(\left({x}^{\mathrm{2}} +\mathrm{2}{xy}^{\mathrm{2}} \right){dx}+\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{1}\right){dy}\right) \\ $$$${where}\:{C}\:{is}\:{the}\:{boundary}\:{of} \\ $$$${region}\:{define}\:{by}\:{y}^{\mathrm{2}} =\:\mathrm{4}{x}\:{and}\:{y} \\ $$$$=\mathrm{1}\:? \\ $$ Answered…

let-f-a-0-pi-d-a-sin-2-a-from-R-1-find-f-a-2-calculate-g-a-0-pi-d-a-sin-2-2-3-calculate-0-pi-d-1-sin-2-and-0-pi-d-2-sin-2-4-calculate-0

Question Number 38210 by prof Abdo imad last updated on 22/Jun/18 $${let}\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{{a}\:+{sin}^{\mathrm{2}} \theta}\:\:\:\left({a}\:{from}\:{R}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{g}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}}…