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 ln | csc x + cot x | + C and ln | csc x cot x | + C equal?
I know both of these are the answer to csc x   d x, and I am able to work them out with calculus using the formulas:
csc x   d x
= csc x csc x cot x csc x cot x   d x
and:
= csc x csc x + cot x csc x + cot x   d x
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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Answers (2)

Kole Weber
Answered 2022-08-14 Author has 16 answers
csc 2 x cot 2 x = 1
Add both terms of ln and use ln 1 = 0
( ln | csc x + cot x | ) + ( ln | csc x cot x | ) = ( ln | csc 2 x cot 2 x | ) = ln 1 = 0 ln | csc x + cot x | = ln | csc x cot x |
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badlife18va
Answered 2022-08-15 Author has 2 answers
ln | csc x cot x | = ln | 1 sin x cos x sin x | = ln | 1 cos x sin x | = ln | sin x 1 cos x | = ln | sin x 1 cos x 1 + cos x 1 + cos x | = ln | sin x 1 cos 2 x ( 1 + cos x ) | = ln | 1 sin x ( 1 + cos x ) | = ln | csc x + cot x | .
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