Question #103214

Integrate tan³x dx

Expert's answer

∫tan⁡3x dx=\int \tan^3x\ dx =


∫sin⁡2x sin⁡xcos⁡3x dx=\int \cfrac{\sin^2x\ \sin x}{\cos^3x}\ dx =


∫1−cos⁡2xcos⁡3xsin⁡x dx=\int \cfrac{1 - \cos^2x}{\cos^3x} \sin x\ dx =


∫cos⁡2x−1cos⁡3x d(cos⁡x)=\int \cfrac{\cos^2x - 1}{\cos^3x}\ d(\cos x) =


∫d(cos⁡x)cos⁡x−∫d(cos⁡x)cos⁡3x=\int \cfrac{d(\cos x)}{\cos x} - \int \cfrac{d(\cos x)}{\cos^3x} =


ln⁡∣cos⁡x∣+12cos⁡2x+C\ln|\cos x| + \cfrac1{2\cos^2x} + C





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