 # Algebraically, how are −ln|cscx+cotx|+C and ln|cscx−cotx|+C equal? I know both of these are the answer to ∫cscx dx, and I am able to work them out with calculus using the formulas: int cscx dx =int csc x (csc x - cot x )/( csc x - cot x) dx and =int csc x (csc x + cot x )/( csc x + cot x) dx Still, when looking at the results, −ln|cscx+cotx|+C and ln|cscx−cotx|+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. wendi1019gt 2022-08-13 Answered
Algebraically, how are $-\mathrm{ln}|\mathrm{csc}x+\mathrm{cot}x|+C$ and $\mathrm{ln}|\mathrm{csc}x-\mathrm{cot}x|+C$ equal?
I know both of these are the answer to , and I am able to work them out with calculus using the formulas:

and:

Still, when looking at the results, $-\mathrm{ln}|\mathrm{csc}x+\mathrm{cot}x|+C$ and $\mathrm{ln}|\mathrm{csc}x-\mathrm{cot}x|+C$ , 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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${\mathrm{csc}}^{2}x-{\mathrm{cot}}^{2}x=1$
Add both terms of ln and use $\mathrm{ln}1=0$
$\left(\mathrm{ln}|\mathrm{csc}x+\mathrm{cot}x|\right)+\left(\mathrm{ln}|\mathrm{csc}x-\mathrm{cot}x|\right)=\left(\mathrm{ln}|{\mathrm{csc}}^{2}x-{\mathrm{cot}}^{2}x|\right)=\mathrm{ln}1=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}|\mathrm{csc}x+\mathrm{cot}x|=-\mathrm{ln}|\mathrm{csc}x-\mathrm{cot}x|$
###### Not exactly what you’re looking for? badlife18va
$\begin{array}{rl}\mathrm{ln}|\mathrm{csc}x-\mathrm{cot}x|& =\mathrm{ln}|\frac{1}{\mathrm{sin}x}-\frac{\mathrm{cos}x}{\mathrm{sin}x}|\\ & =\mathrm{ln}|\frac{1-\mathrm{cos}x}{\mathrm{sin}x}|\\ & =-\mathrm{ln}|\frac{\mathrm{sin}x}{1-\mathrm{cos}x}|\\ & =-\mathrm{ln}|\frac{\mathrm{sin}x}{1-\mathrm{cos}x}\cdot \frac{1+\mathrm{cos}x}{1+\mathrm{cos}x}|\\ & =-\mathrm{ln}|\frac{\mathrm{sin}x}{1-{\mathrm{cos}}^{2}x}\cdot \left(1+\mathrm{cos}x\right)|\\ & =-\mathrm{ln}|\frac{1}{\mathrm{sin}x}\cdot \left(1+\mathrm{cos}x\right)|\\ & =-\mathrm{ln}|\mathrm{csc}x+\mathrm{cot}x|.\end{array}$