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e-x-2-sin-2x-1-cos-2x-dx-

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Question Number 124371 by bemath last updated on 02/Dec/20
∫ ((e^x (2−sin 2x))/(1−cos 2x)) dx
$$\:\int\:\frac{{e}^{{x}} \left(\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}\right)}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}\:{dx}\: \\ $$
Commented by Dwaipayan Shikari last updated on 02/Dec/20
∫e^x ((1/(sin^2 x))−((cosx)/(sinx)))dx =−e^x ((cosx)/(sinx))=−e^x cotx+C
$$\int{e}^{{x}} \left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {x}}−\frac{{cosx}}{{sinx}}\right){dx} \\ $$$$=−{e}^{{x}} \frac{{cosx}}{{sinx}}=−{e}^{{x}} {cotx}+{C} \\ $$
Answered by liberty last updated on 02/Dec/20
∫ ((2e^x )/(1−cos 2x)) −((e^x sin 2x)/(1−cos 2x)) dx = ∫ ((2e^x )/(2sin ^2 x)) − ((2e^x sin xcos x)/(2sin^2 x)) dx = ∫ (e^x /(sin^2 x)) − ((e^x cos x)/(sin x)) dx = ∫ e^x (cosec^2 x−cot x) dx = ∫ (d/dx) (−e^x cot x) dx = −∫ d(e^x cot x) = −e^x cot x + c .
$$\int\:\frac{\mathrm{2}{e}^{{x}} }{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}\:−\frac{{e}^{{x}} \mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}\:{dx}\:= \\ $$$$\int\:\frac{\mathrm{2}{e}^{{x}} }{\mathrm{2sin}\:\:^{\mathrm{2}} {x}}\:−\:\frac{\mathrm{2}{e}^{{x}} \mathrm{sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{2sin}\:^{\mathrm{2}} {x}}\:{dx}\:= \\ $$$$\int\:\frac{{e}^{{x}} }{\mathrm{sin}\:^{\mathrm{2}} {x}}\:−\:\frac{{e}^{{x}} \mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:{dx}\:=\: \\ $$$$\int\:{e}^{{x}} \left(\mathrm{cosec}\:^{\mathrm{2}} {x}−\mathrm{cot}\:{x}\right)\:{dx}\:= \\ $$$$\int\:\frac{{d}}{{dx}}\:\left(−{e}^{{x}} \mathrm{cot}\:{x}\right)\:{dx}\:=\:−\int\:{d}\left({e}^{{x}} \mathrm{cot}\:{x}\right)\: \\ $$$$=\:−{e}^{{x}} \:\mathrm{cot}\:{x}\:+\:{c}\:. \\ $$
Answered by Ar Brandon last updated on 03/Dec/20
I=∫((e^x (2−sin2x))/(1−cos2x))dx=∫e^x {((2−2sinxcosx)/(2sin^2 x))}dx =∫e^x {(1/(sin^2 x))−((cosx)/(sinx))}dx=∫(e^x /(sin^2 x))dx−∫e^x ∙((cosx)/(sinx))dx =−e^x ∙((cosx)/(sinx))+∫e^x ∙((cosx)/(sinx))dx−∫e^x ∙((cosx)/(sinx))dx =−e^x ∙((cosx)/(sinx))+C=−e^x cotx+C
$$\mathcal{I}=\int\frac{\mathrm{e}^{\mathrm{x}} \left(\mathrm{2}−\mathrm{sin2x}\right)}{\mathrm{1}−\mathrm{cos2x}}\mathrm{dx}=\int\mathrm{e}^{\mathrm{x}} \left\{\frac{\mathrm{2}−\mathrm{2sinxcosx}}{\mathrm{2sin}^{\mathrm{2}} \mathrm{x}}\right\}\mathrm{dx} \\ $$$$\:\:\:=\int\mathrm{e}^{\mathrm{x}} \left\{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}−\frac{\mathrm{cosx}}{\mathrm{sinx}}\right\}\mathrm{dx}=\int\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}−\int\mathrm{e}^{\mathrm{x}} \centerdot\frac{\mathrm{cosx}}{\mathrm{sinx}}\mathrm{dx} \\ $$$$\:\:\:=−\mathrm{e}^{\mathrm{x}} \centerdot\frac{\mathrm{cosx}}{\mathrm{sinx}}+\int\mathrm{e}^{\mathrm{x}} \centerdot\frac{\mathrm{cosx}}{\mathrm{sinx}}\mathrm{dx}−\int\mathrm{e}^{\mathrm{x}} \centerdot\frac{\mathrm{cosx}}{\mathrm{sinx}}\mathrm{dx} \\ $$$$\:\:\:=−\mathrm{e}^{\mathrm{x}} \centerdot\frac{\mathrm{cosx}}{\mathrm{sinx}}+\mathcal{C}=−\mathrm{e}^{\mathrm{x}} \mathrm{cotx}+\mathcal{C} \\ $$

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