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Math Help - different results

  1. #1
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    different results

    When integrating (sec x)^2 / (1 - tan x) I obtained ln(mod(1-tan x)) for this using the substitution u=tan x. Why does Wolfram obtain the same result but it is negative? Why does it not use the modulus function? The final result for restricted x values I don't understand.

    integrate[(sec x)^2 / (1 - tan x)] - Wolfram|Alpha
    Last edited by mr fantastic; October 20th 2011 at 04:32 AM. Reason: Included the actual integral.
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  2. #2
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    Re: different results

    Quote Originally Posted by Stuck Man View Post
    I obtained ln(mod(1-tan x)) for this using the substitution u=tan x. Why does Wolfram obtain the same result but it is negative? Why does it not use the modulus function? The final result for restricted x values I don't understand.

    integrate[(sec x)^2 / (1 - tan x)] - Wolfram|Alpha
    It might be because you're supposed to make the substitution u = 1 - tan(x)...
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  3. #3
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    Re: different results

    Why? Is there not many substitutions that could be chosen? Trigonometrical substitution I thought was always done as I did it with u=tan x or sin x or cos x where appropriate.
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  4. #4
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    Re: different results

    Sorry - it doesn't involve trigonometrical substitution.
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  5. #5
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    Re: different results

    u=-tanx would work. It looks like the coefficient must be included which is -1.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: different results

    Quote Originally Posted by Stuck Man View Post
    When integrating (sec x)^2 / (1 - tan x) I obtained ln(mod(1-tan x)) for this using the substitution u=tan x. Why does Wolfram obtain the same result but it is negative? Why does it not use the modulus function? The final result for restricted x values I don't understand.

    integrate[(sec x)^2 / (1 - tan x)] - Wolfram|Alpha
    Omitting the 'absolute value' and the 'arbitrary constant' is...

    \int \frac{\sec^{2} x}{1-\tan x}\ dx = - ln (1-\tan x) = \ln \frac{\cos x}{\cos x - \sin x} =

    = \ln \cos x - \ln (\cos x - \sin x)

    Kind regards

    \chi \sigma
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