Originally Posted by

**scherz0** Hello everyone,

I am trying to prove the following trigonometric identity, but I am not getting the answer given by my textbook:

$\displaystyle -\ln \left|\csc x - \cot x|\right \overset{?}{\mathop{=}}\ -\ln(\sin x) + \ln(\cos x + 1), 0 < x < \pi \text{ } (*)$

Could somoene please comment on my work?

Thank you very much.

*This question actually comes from my attempt to integrate $\displaystyle \int \left -\csc x \right dx $, where $\displaystyle \int \left -\csc x \right dx = -\ln \left|\csc x - \cot x|\right $. I arrived at the LHS of (*) and am trying to get the RHS.

----------------

**My work:**

$\displaystyle LHS = -\ln|\csc x - \cot x| = -\ln \left|\frac{1 - \cos x}{\sin x} \right| $

$\displaystyle = \ln \left |\frac{\sin x}{1 - \cos x} \right| = \ln(\sin x) - \ln|1 - \cos x|$.

1. The last expression doesn't match the textbook's given answer.

2. Also, could someone please justify the removal of the absolute value sign for the second term on the RHS $\displaystyle \left(ie \ln|\cos x + 1| \left \overset{?}{\mathop{=}}\ \right \ln(\cos x + 1) \left \forall \right 0 < x < \pi \right) $?

I know that $\displaystyle \sin x > 0 \left \forall \right 0 < x < \pi$ so I can remove the absolute sign for this term, but $\displaystyle \cos x$ is both negative and positive on this interval.