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Category: Integration

let-f-x-0-1-dt-x-ch-t-1-find-a-explicite-form-of-f-x-2-calculate-0-1-dt-x-ch-t-2-3-find-the-value-of-0-1-dt-1-ch-t-and-0-1-dt-2-ch-t-4

Question Number 42487 by maxmathsup by imad last updated on 26/Aug/18 $${let}\:\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{dt}}{{x}\:+{ch}\left({t}\right)} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicite}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({x}+{ch}\left({t}\right)\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{ch}\left({t}\right)}\:{and}\:\:\int_{\mathrm{0}}…

Question-173523

Question Number 173523 by Physicien last updated on 13/Jul/22 Answered by Mathspace last updated on 13/Jul/22 $${I}=\int\:\:\frac{\mathrm{1}+\sqrt{{x}}−\sqrt{\mathrm{1}+{x}}}{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} −\mathrm{1}−{x}}{dx} \\ $$$$=\int\:\:\frac{\mathrm{1}+\sqrt{{x}}−\sqrt{\mathrm{1}+{x}}}{\mathrm{1}+\mathrm{2}\sqrt{{x}}+{x}−\mathrm{1}−{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{1}+\sqrt{{x}}}{\:\sqrt{{x}}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{{x}}}{dx} \\ $$$${we}\:{have}\:\int\frac{\mathrm{1}+\sqrt{{x}}}{\:\sqrt{{x}}}{dx}=\int\frac{{dx}}{\:\sqrt{{x}}}\:+\int{dx} \\…

BeMath-x-x-2a-x-dx-

Question Number 107965 by bemath last updated on 13/Aug/20 $$\:\:\:\:\frac{\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{} \\ $$$$\int\:{x}\:\sqrt{\frac{{x}}{\mathrm{2}{a}−{x}}}\:{dx}\:?\: \\ $$ Answered by bobhans last updated on 13/Aug/20 $$\:\:\frac{\ddots\mathscr{BOBHANS}\iddots}{\Pi} \\ $$$$\:\mathrm{I}=\:\int\:\mathrm{x}\:\sqrt{\frac{\mathrm{x}}{\mathrm{2a}−\mathrm{x}}}\:\mathrm{dx}\:\:\left[\:\mathrm{set}\:\mathrm{x}\:=\:\mathrm{a}−\mathrm{acos}\:\mathrm{2t}\:\right] \\…

0-cos-ax-sin-bx-x-2-dx-0-1-2sin-2-ax-2-1-2sin-2-bx-2-x-2-dx-2-0-sin-bx-2-x-2-dx-1-2-0-

Question Number 173486 by mnjuly1970 last updated on 12/Jul/22 $$ \\ $$$$\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\:\left({ax}\right)−\:{sin}\left({bx}\right)}{{x}^{\:\mathrm{2}} }{dx}\:\:=? \\ $$$$\:\:\: \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{ax}}{\mathrm{2}}\right)−\left(\mathrm{1}−\mathrm{2}{sin}^{\:\mathrm{2}} \left(\frac{{bx}}{\mathrm{2}}\right)\right)}{{x}\:^{\mathrm{2}} }{dx} \\…

1-1-tanx-dx-

Question Number 42407 by Tawa1 last updated on 25/Aug/18 $$\int\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{tanx}}\:\mathrm{dx} \\ $$ Commented by maxmathsup by imad last updated on 25/Aug/18 $${let}\:{I}\:\:=\:\int\:\:\:\:\:\frac{{dx}}{\mathrm{1}+{tanx}}\:\:{changement}\:{tanx}\:={t}\:{give} \\ $$$${I}\:\:=\:\int\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\:\:{let}\:{decompose}\:{F}\left({t}\right)\:=\:\:\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}}…