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I have reduced the following trig identity to the following which is correct.

$$\int \cos^2(x)\tan^3(x)dx = \int \tan(x) - \sin(x)\cos(x)dx$$

However this next step changes the value of my equation.. i.e False

$$\int \cos^2(x)\tan^3(x)dx = \int \tan(x) dx - \int\sin(x)\cos(x)dx$$

Why can i not do this?

proof below http://www.wolframalpha.com/input/?i=integral%28cos^2x*tan^3x%29+%3D+integral%28tanx%29+-+%28sinxcosx%29

http://www.wolframalpha.com/input/?i=integral%28cos%5E2x%2Atan%5E3x%29+%3D+integral%28tanx%29+-+integral+%28sinxcosx%29

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    $\begingroup$ What? Wolfram gave the exact same result for both. $\endgroup$ – mattos Feb 25 '15 at 12:45
  • $\begingroup$ oops, both links were pointing to the same page $\endgroup$ – Arden Feb 25 '15 at 13:57
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    $\begingroup$ I can't follow the first link - but to comment generally, differences in the apparent form of indefinite integrals are quite often down to hidden differences in the constant of integration. $\endgroup$ – Mark Bennet Feb 25 '15 at 14:01
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The two integrals that Wolfram gets are not equal (so the equation is false), but the two integrals do differ by a constant, so both are correct indefinite integrals.

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  • $\begingroup$ i do not understand your answer completley, i have used similar properties where $$\int x−x^2 dx = \int x dx − \int x^2 dx$$ why is this property not working in this example? Are you suggesting because one willl have 2 constanst while the other wil only have one? $\endgroup$ – Arden Feb 25 '15 at 15:44
  • $\begingroup$ There is a ${}+C$ for an indefinite integral. Mathematica chose its constants differently in the two cases. $\endgroup$ – GEdgar Feb 25 '15 at 18:52

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