Algebraically, how are $-\ln|\csc x + \cot x| +C $ and $\ln| \csc x - \cot x|+C$ equal?
I know both of these are the answer to $\int \csc x \space dx$, and I am able to work them out with calculus using the formulas:
$$\int \csc x \space dx$$ $$=\int \csc x {\csc x - \cot x \over \csc x - \cot x} \space dx$$ and: $$=\int \csc x {\csc x + \cot x \over \csc x + \cot x} \space dx$$
Still, when looking at the results, $-\ln|\csc x + \cot x|+C$ and $\ln| \csc x - \cot x|+C$ , I don't see how these are algebraically equivalent. Perhaps I'm just unaware of some algebra rule (that is likely!). I tried using the Laws of Logs and that doesn't help. Or maybe I'm missing some trig trick.
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$\begingroup$$$\csc^2x-\cot^2x=1$$ Add both terms of ln and use $\ln 1=0$ $$(\ln|\csc x+\cot x|)+(\ln|\csc x-\cot x|)=(\ln|\csc^2x-\cot^2x|)=\ln1=0\\\implies\ln|\csc x+\cot x|=-\ln|\csc x-\cot x|$$
$\endgroup$ 1 $\begingroup$\begin{align} \ln\left|\csc x-\cot x\right|&=\ln\left|\frac1{\sin x}-\frac{\cos x}{\sin x}\right|\\ &=\ln\left|\frac{1-\cos x}{\sin x}\right|\\ &=-\ln\left|\frac{\sin x}{1-\cos x}\right|\\ &=-\ln\left|\frac{\sin x}{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x}\right|\\ &=-\ln\left|\frac{\sin x}{1-\cos^2 x}\cdot(1+\cos x)\right|\\ &=-\ln\left|\frac{1}{\sin x}\cdot(1+\cos x)\right|\\ &=-\ln\left|\csc x+\cot x\right|. \end{align}
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