Question #103214
Integrate tan³x dx
1
Expert's answer
2020-02-17T12:05:23-0500

tan3x dx=\int \tan^3x\ dx =


sin2x sinxcos3x dx=\int \cfrac{\sin^2x\ \sin x}{\cos^3x}\ dx =


1cos2xcos3xsinx dx=\int \cfrac{1 - \cos^2x}{\cos^3x} \sin x\ dx =


cos2x1cos3x d(cosx)=\int \cfrac{\cos^2x - 1}{\cos^3x}\ d(\cos x) =


d(cosx)cosxd(cosx)cos3x=\int \cfrac{d(\cos x)}{\cos x} - \int \cfrac{d(\cos x)}{\cos^3x} =


lncosx+12cos2x+C\ln|\cos x| + \cfrac1{2\cos^2x} + C





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Assignment Expert
17.02.20, 19:31

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17.02.20, 19:28

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