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Question-79413




Question Number 79413 by rajesh4661kumar@gmail.com last updated on 25/Jan/20
Commented by john santu last updated on 25/Jan/20
(((cos xsin x)/(cos 2x(cos^2 x−sin^2 x) )))^2 =  (1/4).((sin^2 2x)/(cos^4 2x)) =(1/4)tan^2 2xsec^2 2x  ∫ (1/8)tan^2 (2x) d(tan(2x)) =  (1/(24))tan^3 (2x) +c
$$\left(\frac{\mathrm{cos}\:{x}\mathrm{sin}\:{x}}{\mathrm{cos}\:\mathrm{2}{x}\left(\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)\:}\right)^{\mathrm{2}} = \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{2}{x}}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{2}} \mathrm{2}{x}\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:{d}\left(\mathrm{tan}\left(\mathrm{2}{x}\right)\right)\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{24}}\mathrm{tan}\:^{\mathrm{3}} \left(\mathrm{2}{x}\right)\:+{c}\: \\ $$

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