Equivalent forms for $\int \csc(x) dx$

Question

This question asked about two common forms commonly given for $\int \csc(x) dx$, specifically:

$$ (1) \qquad \int \csc(x) dx = \mathbf{{\color{red}-} \ ln|csc(x)+cot(x)| + C}, $$

and

$$ (2) \qquad \int \csc(x) dx = \mathbf{ln|csc(x) \ {\color{red}-}\ cot(x)| + C}. $$

And the accepted answer is clear and convincing; however, WolframAlpha seems to disagree.

In the past, I've found that trying to understand why Wolfram is giving an answer that is different from what I'd expect generally leads me to learn something new, so I was hoping someone might be able to offer some insight on this one.

Answer 1

Essentially all we need to show is that $\exists C$ such that: $$-\ln |\csc x + \cot x| = \ln |\csc x - \cot x| + C$$

This is equivalent to showing that:

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

Well ... we can actually just show that $\frac{1}{\csc x + \cot x} = \csc x - \cot x$.

As usual when dealing with trigonometric functions, things become a lot more clear when we convert everything to $\cos x$ and $\sin x$ instead. The left-hand side becomes:

$$\frac{1}{\frac{1}{\sin x} + \frac{\cos x}{\sin x}} = \frac{\sin x}{1 + \cos x}$$

and the right-hand side is:

$$\frac{1 - \cos x}{\sin x}$$

We can bring both sides to the same denominator:

$$ \begin{cases} LHS &=& \frac{\sin^2 x}{\sin x (1 + \cos x)} \\ RHS &=& \frac{1 - \cos^2 x}{\sin x(1 + \cos x)} \end{cases} $$

It is easy to see that these are equal.

Answer 2

Well: $$\frac{1}{\sin x}+\frac{1}{\tan x}=\frac{1+\cos x}{\sin x}\cdot\frac{1-\cos x}{1-\cos x}=\frac{1-\cos^2x}{\sin x(1-\cos x)}$$

$$=\frac{\sin^2 x}{\sin x(1-\cos x)}=\frac{\sin x}{1-\cos x}=\bigg(\frac{1}{\sin x}-\frac{1}{\tan x}\bigg)^{-1} \ a.r.$$ so the identity is clearly true, wherever $\csc x,\cot x$ are defined.

My first suspicion was that maybe "inequality" exists when the functions aren't defined, for example at $x=0$. But, checking this threw that out the window. It seems like an error to me.