Absolute value in integral

  • Thread starter MathewsMD
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  • #1
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∫tan(x) dx = -ln lcos(x)l + C = f(x)

So is f'(x) = -sin(x)/lcos(x)l ? When taking the derivative, do we only consider f'(x) to only exist on the intervals where cos(x) > 0 instead?
 

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Dick
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∫tan(x) dx = -ln lcos(x)l + C = f(x)

So is f'(x) = -sin(x)/lcos(x)l ? When taking the derivative, do we only consider f'(x) to only exist on the intervals where cos(x) > 0 instead?

No, the derivative of log(|y(x)|) is the derivative of log(y(x)) if y(x)>0 and it's the derivative of log(-y(x)) if y(x)<0. Can you show that they are both (1/y(x))*y'(x). Not (1/|y(x)|)*y'(x)?
 

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