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.
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$You can also find the answer that wolfram alpha provides by repeating almost exactly the work done in the original post, but instead multiplying and dividing by $\csc x+\cot x$. Even the same $u$-substitution works except the whole numerator is off by a minus sign, which leads to the extra negative outside the logarithm in the antiderivative.
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