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




Question Number 124043 by liberty last updated on 30/Nov/20
 ∫ ((sin x cos x)/(1−2cos 2x)) dx
$$\:\int\:\frac{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{2cos}\:\mathrm{2}{x}}\:{dx}\: \\ $$
Answered by john_santu last updated on 30/Nov/20
Answered by Dwaipayan Shikari last updated on 30/Nov/20
∫((sinxcosx)/(1−2cos2x))dx       1−2cos2x=t⇒4sin2x=(dt/dx)  =(1/8)∫(dt/t) =(1/8)log(1−2cos2x)+C
$$\int\frac{{sinxcosx}}{\mathrm{1}−\mathrm{2}{cos}\mathrm{2}{x}}{dx}\:\:\:\:\:\:\:\mathrm{1}−\mathrm{2}{cos}\mathrm{2}{x}={t}\Rightarrow\mathrm{4}{sin}\mathrm{2}{x}=\frac{{dt}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{dt}}{{t}}\:=\frac{\mathrm{1}}{\mathrm{8}}{log}\left(\mathrm{1}−\mathrm{2}{cos}\mathrm{2}{x}\right)+{C} \\ $$

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