Integration formula of cosecant

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In summary, the conversation discussed the integration formula of cosecant, which can be written as either "= ln |csc x - cot x| + C" or "= - ln |csc x + cot x| + C". The difference in notation is due to the fact that both equations are correct, as shown through differentiation and simplification. The only difference is the positive or negative result when the absolute value is removed.
  • #1
TGV320
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TL;DR Summary
Small differences between formulas
Hi

I have a question about the integration formula of cosecant which leaves me puzzled.

I usually find it written as " = ln |csc x - cot x| + C" in most manuals, but sometimes it is written as "= - ln |csc x + cot x| + C" or "= - ln (csc x + cot x) + C".

Why is that? Can they all be used?

Thanks a lot
 
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  • #2
TGV320 said:
TL;DR Summary: Small differences between formulas

Hi

I have a question about the integration formula of cosecant which leaves me puzzled.

I usually find it written as " = ln |csc x - cot x| + C" in most manuals, but sometimes it is written as "= - ln |csc x + cot x| + C" or "= - ln (csc x + cot x) + C".

Why is that? Can they all be used?

Thanks a lot
Have you tried differentiating each one to check that they are all anti-derivatives of ##cosec##? Sometimes functions that look different only differ by a constant. E.g:
$$\cos^2 x = 1 - \sin^2x$$Which means that:$$\frac d {dx} \cos^2 x = - \frac d {dx} sin^2 x$$Check that out if you want.
 
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For both expressions for the antiderivative to be correct, we must have [tex]
\ln |\csc x - \cot x| + \ln |\csc x + \cot x| = \ln |\csc^2 x - \cot^2 x| = 0.[/tex] So can we show that [itex]|\csc^2 x - \cot^2 x| = 1[/itex]?
 
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  • #4
Hi

Since cscx^2-cotx^2=1，I think it is true then, both equations do work indeed.
I have also tried to differentiate the results, and it seems that the only thing that varies is what comes out of the absolute value, therefore conditioning the positive of negative of the result.

Thanks a lot， I am grateful for your help
 

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