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I understand that for solving the integral

$\int\sec^3(x)\sin(x) dx$ can be solved using $\tan(x)\sec^2(x) dx$

but why cant I solve this integral using U substitution?

$$\frac{1}{\cos^3(x)\sin(x)}$$

let $$ u =\cos(x)\quad du= -\sin(x) dx \\ \quad \\ -du =\sin(x) dx \\ \int\frac{1}{u^3}\cdot(-du) \\ ∫-u^{-3} du \\ -u^{-2}/-2 = \\ \frac{\cos^{-2}(x)}{2}$$

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    $\begingroup$ I'm finding it hard to understand your question. Please format it using MathJax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – mrnovice Apr 7 '17 at 22:45
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    $\begingroup$ You can. But don 't forget your "+C". $\frac 12 \tan^2 x + c = \frac 12 \sec^2 x + c = \frac 12 \cos^{-2} + c$ $\endgroup$ – Doug M Apr 7 '17 at 22:46
  • $\begingroup$ You can.. you answered it correctly.. $\endgroup$ – Rab Apr 7 '17 at 22:46
  • $\begingroup$ And you've garbled at least two minus signs that cancelled out. $\endgroup$ – Chappers Apr 7 '17 at 22:47
  • $\begingroup$ For basic information about writing math on MSE see here, here, here and here. $\endgroup$ – user409521 Apr 7 '17 at 22:59
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This is actually a correct way to go about it and your answer is fine (except for the missing $+C$ constant, but we'll let it slide).

The only reason not to take this answer as valid is if your professor specifically asked for the other version, otherwise this method should be fair game

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