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I have to prove that:

$x \sec x - \ln |\sec x + \tan x| + C$

is the indefinite integral of:

$x \sec x \tan x $

by taking the derivative.

I've got far enough to get:

$x\sec x\tan x + \sec x -\dfrac{|\sec x+\tan x|(\sec^2 x + \sec x \tan x)}{|\sec x + \tan x|}.$

Kind of stuck here. Am I able to cancel out the $|\sec x + \tan x|$ on top and bottom and then set $-\sec x$ equal to $\sec^2 x + \sec x \tan x$? I'm guessing that's not right though.

Sorry for the crummy way I have it setup, feel free to edit it.

2 Answers 2

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Two issues—first, as suggested in Jerry's answer, you have a factor of $|\sec x+\tan x|$ in the numerator of the last term of your derivative that does not belong there. Second, the derivative of $\ln|x|$ is $\frac{1}{x}$ (no absolute value), so $\frac{d}{dx}\ln|f(x)|=\frac{1}{f(x)}\cdot f'(x)$ with the chain rule.

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    Yes I'm not sure what I did but it looks funky, the way you wrote it makes way more sense. Thanks man.2011-02-08
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Your derivative for $x\sec\;x$ is correct; for the second term, note that $\frac{\mathrm d}{\mathrm dx}\ln(f(x))=\frac{f^{\prime}(x)}{f(x)}$ . Apply the formula accordingly.

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    @Finzz: Yes, you did -- you wrote $f\cdot f'/f$ instead of $f'/f$.2011-02-08