# Clarification

• Mar 28th 2010, 06:33 PM
kaiser0792
Clarification
This is not strictly a calculus problem, but a question about the form of a trigonometric integral.

Can someone tell me how -ln|csc x + cot x| is equal to ln|csc x - cot x|?

I know that using the properties of logarithms that

-ln|csc x + cot x| = ln (|csc x + cot x|)^-1 = ln |[1/(csc x + cot x)]|, but, I'm drawing a blank as to how any of these forms translates to
ln|csc x - cot x|.
• Mar 28th 2010, 06:39 PM
Prove It
Quote:

Originally Posted by kaiser0792
This is not strictly a calculus problem, but a question about the form of a trigonometric integral.

Can someone tell me how -ln|csc x + cot x| is equal to ln|csc x - cot x|?

I know that using the properties of logarithms that

-ln|csc x + cot x| = ln (|csc x + cot x|)^-1 = ln |[1/(csc x + cot x)]|, but, I'm drawing a blank as to how any of these forms translates to
ln|csc x - cot x|.

$|\csc{x} + \cot{x}|^{-1} = \left|\frac{1}{\csc{x} + \cot{x}}\right|$

$= \left|\frac{\csc{x} - \cot{x}}{(\csc{x} + \cot{x})(\csc{x} - \cot{x})}\right|$

$= \left|\frac{\csc{x} - \cot{x}}{\csc^2{x} - \cot^2{x}}\right|$

But $\csc^2{x} - \cot^2{x} = 1$

So $\left|\frac{\csc{x} - \cot{x}}{\csc^2{x} - \cot^2{x}}\right| = \left|\frac{\csc{x} - \cot{x}}{1}\right|$

$= |\csc{x} - \cot{x}|$.
• Mar 28th 2010, 06:56 PM
kaiser0792
Thank you Prove IT, I knew it was an algebra step, but I've been deep into trig integrals and couldn't see the forest for the trees!