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Question Number 5800 by sumit last updated on 28/May/16

∫((cos2x)/(sin^2 x∙cos^2 x))dx

$$\int\frac{{cos}\mathrm{2}{x}}{{sin}^{\mathrm{2}} {x}\centerdot{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$ \\ $$

Commented by Boma last updated on 09/Aug/19

Boma

Answered by Yozzii last updated on 28/May/16

∫((cos2x)/(sin^2 xcos^2 x))dx  =∫((cos2x)/((((sin2x)/2))^2 ))dx  =4∫((cos2x)/(sin^2 2x))dx  =2∫(1/u^2 )du      (u=sin2x)  =−(2/u)+c  ∫((cos2x)/(sin^2 xcos^2 x))dx=c−(2/(sin2x))

$$\int\frac{{cos}\mathrm{2}{x}}{{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int\frac{{cos}\mathrm{2}{x}}{\left(\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}\int\frac{{cos}\mathrm{2}{x}}{{sin}^{\mathrm{2}} \mathrm{2}{x}}{dx} \\ $$$$=\mathrm{2}\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} }{du}\:\:\:\:\:\:\left({u}={sin}\mathrm{2}{x}\right) \\ $$$$=−\frac{\mathrm{2}}{{u}}+{c} \\ $$$$\int\frac{{cos}\mathrm{2}{x}}{{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}{dx}={c}−\frac{\mathrm{2}}{{sin}\mathrm{2}{x}} \\ $$

Commented by Boma last updated on 09/Aug/19

Boma

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