1. ## 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&#91;&#40;sec x&#41;&#94;2 &#47; &#40;1 - tan x&#41;&#93; - Wolfram|Alpha

2. ## Re: different results

Originally Posted by Stuck Man
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&#91;&#40;sec x&#41;&#94;2 &#47; &#40;1 - tan x&#41;&#93; - Wolfram|Alpha
It might be because you're supposed to make the substitution u = 1 - tan(x)...

3. ## 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.

4. ## Re: different results

Sorry - it doesn't involve trigonometrical substitution.

5. ## Re: different results

u=-tanx would work. It looks like the coefficient must be included which is -1.

6. ## Re: different results

Originally Posted by Stuck Man
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&#91;&#40;sec x&#41;&#94;2 &#47; &#40;1 - tan x&#41;&#93; - Wolfram|Alpha
Omitting the 'absolute value' and the 'arbitrary constant' is...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$