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Question Number 78581 by john santu last updated on 18/Jan/20

∫ ((2xsin 2x)/((2x−sin 2x)^2 )) dx ?

$$\int\:\frac{\mathrm{2}{x}\mathrm{sin}\:\mathrm{2}{x}}{\left(\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} }\:{dx}\:? \\ $$

Commented by john santu last updated on 19/Jan/20

consider ((2x sin 2x)/((2x−sin 2x)^2 )) ×((1−cos 2x)/(1−cos 2x)) = ((2x sin 2x)/(1−cos 2x)) × ((1−cos 2x)/((2x−sin 2x)^2 )) now use integration by parts where u = ((2xsin 2x)/(1−cos 2x)) ⇒du=((sin 2x−2x)/(1−cos 2x)) dv = ((1−cos 2x)/((2x−sin 2x))) dx ⇒ v = (1/(sin 2x−2x)) I = u.v −∫ v du I = ((2xsin 2x)/(1−cos 2x)). (1/(sin 2x−2x)) − ∫ ((1/(sin 2x−2x))). ((sin 2x−2x)/(1−cos 2x)) dx I = ((2x sin 2x)/((1−cos 2x)(sin 2x−2x)))− ∫ (dx/(1−cos 2x)) I= ((2x sin 2x)/((1−cos 2x)(sin 2x−2x))) −∫ (dx/(2sin^2 x)) I = ((2x sin 2x)/((1−cos 2x)(sin 2x−2x))) +((cot x)/2) + c

$${consider}\: \\ $$$$\frac{\mathrm{2}{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\left(\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} }\:×\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}\:= \\ $$$$\frac{\mathrm{2}{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}\:×\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\left(\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$${now}\:{use}\:{integration}\:{by}\:{parts} \\ $$$${where}\:{u}\:=\:\frac{\mathrm{2}{x}\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}\:\Rightarrow{du}=\frac{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}} \\ $$$${dv}\:=\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\left(\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}\right)}\:{dx}\:\Rightarrow\:{v}\:=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}} \\ $$$${I}\:=\:{u}.{v}\:−\int\:{v}\:{du} \\ $$$${I}\:=\:\frac{\mathrm{2}{x}\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}.\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}}\:−\:\int\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}}\right).\:\frac{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}\:{dx} \\ $$$${I}\:=\:\frac{\mathrm{2}{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)}−\:\int\:\:\frac{{dx}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}} \\ $$$${I}=\:\frac{\mathrm{2}{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)}\:−\int\:\frac{{dx}}{\mathrm{2sin}\:^{\mathrm{2}} {x}} \\ $$$${I}\:=\:\frac{\mathrm{2}{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)}\:+\frac{\mathrm{cot}\:{x}}{\mathrm{2}}\:+\:{c} \\ $$

Commented by john santu last updated on 19/Jan/20

I=((4x sin x cos x)/((2sin^2 x)(sin 2x−2x)))+((cos x)/(2sin x)) I= ((4x cos x)/(2sin x(sin 2x−2x))) + ((cos x)/(2sin x)) I= ((4xcos x +sin 2x cos x −2x cos x)/(2sin x(sin 2x−2x))) I = ((2x cos x+sin 2xcos x)/(2sin x(sin 2x−2x))) +c I= ((cos x(2x+sin 2x))/(2sin x(sin 2x−2x))) +c

$${I}=\frac{\mathrm{4}{x}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\left(\mathrm{2sin}\:^{\mathrm{2}} {x}\right)\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)}+\frac{\mathrm{cos}\:{x}}{\mathrm{2sin}\:{x}} \\ $$$${I}=\:\frac{\mathrm{4}{x}\:\mathrm{cos}\:{x}}{\mathrm{2sin}\:{x}\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)}\:+\:\frac{\mathrm{cos}\:{x}}{\mathrm{2sin}\:{x}} \\ $$$${I}=\:\frac{\mathrm{4}{x}\mathrm{cos}\:{x}\:+\mathrm{sin}\:\mathrm{2}{x}\:\mathrm{cos}\:{x}\:−\mathrm{2}{x}\:\mathrm{cos}\:{x}}{\mathrm{2sin}\:{x}\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)} \\ $$$${I}\:=\:\frac{\mathrm{2}{x}\:\mathrm{cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}\mathrm{cos}\:{x}}{\mathrm{2sin}\:{x}\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)}\:+{c} \\ $$$${I}=\:\frac{\mathrm{cos}\:{x}\left(\mathrm{2}{x}+\mathrm{sin}\:\mathrm{2}{x}\right)}{\mathrm{2sin}\:{x}\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2}{x}\right)}\:+{c} \\ $$

Commented by jagoll last updated on 19/Jan/20

waww

$$\mathrm{waww} \\ $$

Commented by john santu last updated on 19/Jan/20

thanks

$${thanks} \\ $$

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