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Where's the mistake in this solution? $$\int \tan^3x\sec^2xdx = \int \frac{\sin^3x}{\cos^5x}dx=\int\frac{\sin x(1-\cos²x)}{\cos^5x}dx=\int\frac{u^2-1}{u^5}du$$$$=\frac{1}{4u^4}-\frac{1}{2u^2}+C=\frac{1}{4\cos^4x}-\frac{1}{2\cos^2x}+C$$ for $ \cos x=u \to du=-\sin xdx$.

I also tried doing $$\int \tan^3x\sec^2xdx = \int u^3du=\frac{u^4}{4}+C=\frac{\tan^4x}{4}+C$$ for $ tgx=u \to du=\sec^2xdx$


Please help!

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    $\begingroup$ Both your answers are correct; the 2 functions just differ by a constant. $\endgroup$ – user84413 Feb 25 '15 at 19:51
  • $\begingroup$ Have you tried diffentiating your 2 answers to see if one or the other are the right (or wrong) answers? Have you tried subtracting and simplifying your two answers? If you do one or the other of these actions, you will find that both your answers are good (... up to a constant). $\endgroup$ – Bernard Massé Feb 25 '15 at 19:53
  • $\begingroup$ Thanks! I didn't tried it yet. As the exercise's answer was the second one, i tried again thinking that mine was wrong. $\endgroup$ – duvidasnorio Feb 25 '15 at 20:47
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\begin{align} \frac{1}{4\cos^4x}-\frac{1}{2\cos^2x} +\frac{1}{4} &= \frac{1-2\cos^2 x+\cos^4x}{4\cos^4 x}\\ &= \frac{(\cos^2x-1)^2}{4\cos^4x}\\ &= \frac{(-\sin^2x )^2}{4\cos^4x}\\ &= \frac{\tan^4x}{4} \end{align}

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