Menu Close

Category: Integration

calculate-A-0-cos-4-x-x-2-2-2-dx-2-find-the-value-of-0-cos-4-x-x-2-3-2-dx-

Question Number 136406 by mathmax by abdo last updated on 21/Mar/21 $$\mathrm{calculate}\:\:\mathrm{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}}…

find-U-n-1-n-n-1-1-x-2-arctan-x-1-x-dx-and-lim-n-U-n-

Question Number 136402 by mathmax by abdo last updated on 21/Mar/21 $$\mathrm{find}\:\mathrm{U}_{\mathrm{n}} =\int_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\mathrm{n}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{arctan}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx} \\ $$$$\mathrm{and}\:\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \mathrm{U}_{\mathrm{n}} \\ $$ Terms of Service Privacy…

sdvanced-cslculus-if-x-R-and-x-0-x-e-t-1-t-ln-x-t-dt-then-prove-that-0-e-x-x-dx-2-

Question Number 136381 by mnjuly1970 last updated on 21/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:……{sdvanced}\:\:\:{cslculus}…… \\ $$$$\:{if}\:\:{x}\in\mathbb{R}^{+} \:{and}::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} \frac{{e}^{{t}} −\mathrm{1}}{{t}}{ln}\left(\frac{{x}}{{t}}\right){dt} \\ $$$$\:\:{then}\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}} \boldsymbol{\phi}\left({x}\right){dx}=\zeta\left(\mathrm{2}\right)…

dx-sin-x-cos-x-

Question Number 136365 by liberty last updated on 21/Mar/21 $$\int\:\frac{{dx}}{\mathrm{sin}\:{x}\:\sqrt{\mathrm{cos}\:{x}}}\:=? \\ $$ Answered by mathmax by abdo last updated on 21/Mar/21 $$\mathrm{I}\:=\int\:\frac{\mathrm{dx}}{\mathrm{sinx}\sqrt{\mathrm{cosx}}}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{cosx}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{x}=\mathrm{arcos}\left(\mathrm{t}^{\mathrm{2}} \right) \\…