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Category: Integration

Question-142060

Question Number 142060 by iloveisrael last updated on 26/May/21 Answered by Ar Brandon last updated on 26/May/21 $$\mathrm{I}=\int\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{sec}\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \right)\mathrm{tan}\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \right)\mathrm{dx} \\ $$$${u}=\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \Rightarrow\mathrm{d}{u}=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…

dx-e-2x-4-sec-1-e-x-4-

Question Number 76467 by kaivan.ahmadi last updated on 27/Dec/19 $$\int\frac{{dx}}{\:\sqrt{{e}^{\mathrm{2}{x}} −\mathrm{4}}\left({sec}^{−\mathrm{1}} \left(\frac{{e}^{{x}} }{\mathrm{4}}\right)\right)} \\ $$ Answered by john santu last updated on 28/Dec/19 $${let}\::\mathrm{sec}\:{u}\:=\:\frac{{e}^{{x}} }{\mathrm{4}}\:\rightarrow\:{e}^{{x}}…

cothx-dx-

Question Number 76469 by kaivan.ahmadi last updated on 27/Dec/19 $$\int{cothx}\:{dx} \\ $$ Commented by mathmax by abdo last updated on 27/Dec/19 $$\int\:{coth}\left({x}\right){dx}\:=\int\:\frac{{ch}\left({x}\right)}{{sh}\left({x}\right)}{dx}\:=\int\:\:\frac{{e}^{{x}} \:+{e}^{−{x}} }{{e}^{{x}} −{e}^{−{x}}…

0-pi-2-sin-40x-sin-5x-dx-

Question Number 141998 by iloveisrael last updated on 25/May/21 $$\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{sin}\:\left(\mathrm{40}{x}\right)}{\mathrm{sin}\:\left(\mathrm{5}{x}\right)}\:{dx}\: \\ $$ Answered by EDWIN88 last updated on 25/May/21 $$\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{\mathrm{sin}\:\left(\mathrm{40x}\right)}{\mathrm{sin}\:\left(\mathrm{5x}\right)}\:\mathrm{dx}\: \\…

cos3x-e-2x-dx-

Question Number 76462 by kaivan.ahmadi last updated on 27/Dec/19 $$\int{cos}\mathrm{3}{x}\:{e}^{−\mathrm{2}{x}} {dx} \\ $$ Commented by mathmax by abdo last updated on 27/Dec/19 $${let}\:{A}\:=\int\:{cos}\left(\mathrm{3}{x}\right){e}^{−\mathrm{2}{x}} {dx}\:\Rightarrow{A}\:={Re}\left(\int\:{e}^{\mathrm{3}{ix}} \:{e}^{−\mathrm{2}{x}}…

Question-76442

Question Number 76442 by aliesam last updated on 27/Dec/19 Answered by Tanmay chaudhury last updated on 28/Dec/19 $${x}={tana}\:\:\:\:{dx}={sec}^{\mathrm{2}} {ada} \\ $$$${sec}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{tan}^{\mathrm{2}} {a}}{\mathrm{1}−{tan}^{\mathrm{2}} {a}}\right)\rightarrow{sec}^{−\mathrm{1}} \left({sec}\mathrm{2}{a}\right)=\mathrm{2}{a}…