I'm getting a couple of different answers from different sources, so I'd like to verify something.
I misplaced my original notes from my prof, but working from memory I have the following:
\begin{align}
\int\csc(x)\ dx&=\int\csc(x)\left(\frac{\csc(x)-\cot(x)}{\csc(x)-\cot(x)}\right)\ dx\\
&=\int\frac{\csc^{2}(x)-\csc(x)\cot(x)}{\csc(x)-\cot(x)}\ dx\\
&=\int\frac{1}{u}\ du\\
&=\ln|u|+C\\
&=\ln|\csc(x)-\cot(x)|+C
\end{align}
This looks proper when I trace it, but wolfram alpha is saying that the answer should be
$$-\ln|\csc(x)+\cot(x)|+C$$
Sadly, it doesn't provide a step-by-step. It just says that's the answer.
So which is it? Or are they both equivalent? I've never been great with the laws of logarithms.
Best Answer
Start with the identity
$$(\csc x-\cot x)(\csc x+\cot x)=\csc^2x-\cot^2x=1\;;$$
this implies that
$$|\csc x-\cot x|\cdot|\csc x+\cot x|=|\csc^2x-\cot^2x|=1\;.$$
Now use the fact that if $a,b>0$ and $ab=1$, then $\ln a+\ln b=\ln 1=0$, so $\ln a=-\ln b$ to conclude that
$$\ln|\csc x-\cot x|=-\ln|\csc x+\cot x|\;,$$
and the two answers are the same.