Menu Close

Category: Integration

Question-41675

Question Number 41675 by Raj Singh last updated on 11/Aug/18 Commented by math khazana by abdo last updated on 12/Aug/18 $${I}\:=\:\int\:\:\:\:\frac{{dx}}{\mathrm{1}+\frac{{cosx}}{{sinx}}}\:=\:\int\:\:\:\frac{{sinx}}{{sinx}\:+{cosx}}\:{dx} \\ $$$$=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…

calculate-A-0-pi-4-cos-8-xdx-and-B-0-pi-4-sin-8-xdx-2-calculate-A-B-and-A-B-3-calculate-A-2-B-2-

Question Number 41677 by math khazana by abdo last updated on 11/Aug/18 $${calculate}\:{A}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{8}} {xdx}\:{and}\: \\ $$$${B}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}^{\mathrm{8}} {xdx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}\:+{B}\:{and}\:{A}−{B} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{A}^{\mathrm{2}}…

JS-0-pi-6-sin-2-6x-cos-4-3x-dx-by-using-the-Gamma-function-

Question Number 107197 by john santu last updated on 09/Aug/20 $$\:\:\:\:\:\trianglerighteq\mathrm{JS}\trianglelefteq \\ $$$$\int\overset{\:\pi/\mathrm{6}} {\:}_{\mathrm{0}} \mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{6x}\right)\:\mathrm{cos}\:^{\mathrm{4}} \left(\mathrm{3x}\right)\:\mathrm{dx}\:? \\ $$$$\left[\:\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\mathcal{G}\mathrm{amma}\:\mathrm{function}\:\right] \\ $$ Terms of Service Privacy…

1-x-x-2-x-6-dx-

Question Number 172712 by ilhamQ last updated on 30/Jun/22 $$\mathrm{1}.\:\int\frac{{x}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}\:{dx}=… \\ $$ Answered by Ar Brandon last updated on 30/Jun/22 $$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{dx} \\…

Question-41634

Question Number 41634 by Tawa1 last updated on 10/Aug/18 Commented by maxmathsup by imad last updated on 10/Aug/18 $${I}\:\:=\:\int\:\:\:\:\frac{{dx}}{{sinx}\:+\frac{\mathrm{1}}{{cosx}}}\:=\:\:\int\:\:\:\:\:\:\frac{{cosx}}{{sinx}\:.{cosx}\:+\mathrm{1}}{dx}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\:\:\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}}…

Question-41627

Question Number 41627 by alex041103 last updated on 10/Aug/18 Commented by alex041103 last updated on 10/Aug/18 $${Just}\:{for}\:{fun}!\:{I}\:{have}\:{already}\:{posted} \\ $$$${this}\:{Q}.\:{around}\:{a}\:{year}\:{ago}.\:{The}\:{result} \\ $$$${is}\:{beautiful}\:{and}\:{worths}\:{giving}\:{it}\:{a}\:{try}. \\ $$$${Can}\:{you}\:{prove}\:{something}\:{interesting}. \\ $$$${Hint}:\:{Calculate}\:{the}\:{result}.…