# Math Help - Simplification of -ln| cscx - cotx |

1. ## Simplification of -ln| cscx - cotx |

Hello everyone,

I am trying to prove the following trigonometric identity, but I am not getting the answer given by my textbook:

$-\ln \left|\csc x - \cot x|\right \overset{?}{\mathop{=}}\ -\ln(\sin x) + \ln(\cos x + 1), 0 < x < \pi \text{ } (*)$

Could somoene please comment on my work?

Thank you very much.

*This question actually comes from my attempt to integrate $\int \left -\csc x \right dx$, where $\int \left -\csc x \right dx = -\ln \left|\csc x - \cot x|\right$. I arrived at the LHS of (*) and am trying to get the RHS.

----------------

My work:

$LHS = -\ln|\csc x - \cot x| = -\ln \left|\frac{1 - \cos x}{\sin x} \right|$

$= \ln \left |\frac{\sin x}{1 - \cos x} \right| = \ln(\sin x) - \ln|1 - \cos x|$.

1. The last expression doesn't match the textbook's given answer.

2. Also, could someone please justify the removal of the absolute value sign for the second term on the RHS $\left(ie \ln|\cos x + 1| \left \overset{?}{\mathop{=}}\ \right \ln(\cos x + 1) \left \forall \right 0 < x < \pi \right)$?

I know that $\sin x > 0 \left \forall \right 0 < x < \pi$ so I can remove the absolute sign for this term, but $\cos x$ is both negative and positive on this interval.

2. ## Re: Simplification of -ln| cscx - cotx |

hi scherz0

$-\ln{|\frac{\sin{x}}{1-\cos{x}}|}=-(\ln{|\sin{x}|}-ln{|1-\cos{x}|})=-\ln{|\sin{x}|}+ln{|1-\cos{x}|}$

and as for cos(x)+1,that's probably their mistake or a typo.

also cos(x)+1 is always greater than zero because:
-1<=cos(x)<=1
add 1 to everything to get:
0<=cos(x)+1<=2
so you don't need the abolute value sign.

3. ## Re: Simplification of -ln| cscx - cotx |

Hi anonimnystefy,

Thank you for your response.

However, could you please explain why you still have the negative term in:

$-\ln \left | \frac{\sin{x}}{1-\cos{x}} \right |$?

I believe that it should be, as I wrote in my original post:

$-\ln \left | \frac{1-\cos{x}}{\sin{x}} \right | = + \ln \left | \frac{\sin{x}}{1-\cos{x}} \right |$

4. ## Re: Simplification of -ln| cscx - cotx |

hi scherz0

i have edited my post so now it also responds to your second question now.

i am glad i could help.

5. ## Re: Simplification of -ln| cscx - cotx |

Hi anonimnystefy,

Thanks again for your response.

But would you mind looking at post #3? I meant to post this before but you were ahead of me .

Are the signs supposed to be as they are in your post (#2)?

6. ## Re: Simplification of -ln| cscx - cotx |

hi scherz0

did you try differentianting your and the book's answer to see which one is correct?

it is very possible the book got it wrong.it happens to me all the time.

7. ## Re: Simplification of -ln| cscx - cotx |

Originally Posted by anonimnystefy
...

did you try differentiating your and the book's answer to see which one is correct?
...
Yes, differentiating the two results is a good way to check them. Perhaps the two results differ only by a constant, in which case both could be correct.

8. ## Re: Simplification of -ln| cscx - cotx |

Originally Posted by scherz0
*This question actually comes from my attempt to integrate $\int \left -\csc x \right dx$, where $\int \left -\csc x \right dx = -\ln \left|\csc x - \cot x|\right$. I arrived at the LHS of (*) and am trying to get the RHS.
your integral is incorrect ...

$\int -\csc{x} \, dx =$

$\int -\csc{x} \cdot \frac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}} \, dx =$

$\int \frac{-csc^2{x} - \csc{x}\cot{x}}{\csc{x}+\cot{x}} \, dx =$

$\ln|\csc{x}+\cot{x}| + C$

also, note that ...

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

9. ## Re: Simplification of -ln| cscx - cotx |

Originally Posted by scherz0
Hello everyone,

I am trying to prove the following trigonometric identity, but I am not getting the answer given by my textbook:

$-\ln \left|\csc x - \cot x|\right \overset{?}{\mathop{=}}\ -\ln(\sin x) + \ln(\cos x + 1), 0 < x < \pi \text{ } (*)$

Could somoene please comment on my work?

Thank you very much.

*This question actually comes from my attempt to integrate $\int \left -\csc x \right dx$, where $\int \left -\csc x \right dx = -\ln \left|\csc x - \cot x|\right$. I arrived at the LHS of (*) and am trying to get the RHS.

----------------

My work:

$LHS = -\ln|\csc x - \cot x| = -\ln \left|\frac{1 - \cos x}{\sin x} \right|$

$= \ln \left |\frac{\sin x}{1 - \cos x} \right| = \ln(\sin x) - \ln|1 - \cos x|$.

1. The last expression doesn't match the textbook's given answer.

2. Also, could someone please justify the removal of the absolute value sign for the second term on the RHS $\left(ie \ln|\cos x + 1| \left \overset{?}{\mathop{=}}\ \right \ln(\cos x + 1) \left \forall \right 0 < x < \pi \right)$?

I know that $\sin x > 0 \left \forall \right 0 < x < \pi$ so I can remove the absolute sign for this term, but $\cos x$ is both negative and positive on this interval.
Also, what is the justification in removing the absolute value sign from the first term?

10. ## Re: Simplification of -ln| cscx - cotx |

Originally Posted by scherz0
*This question actually comes from my attempt to integrate $\int \left -\csc x \right dx$, where $\int \left -\csc x \right dx = -\ln \left|\csc x - \cot x|\right$. I arrived at the LHS of (*) and am trying to get the RHS.

Originally Posted by skeeter
your integral is incorrect ...

$\int -\csc{x} \, dx =$

$\int -\csc{x} \cdot \frac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}} \, dx =$

$\int \frac{-csc^2{x} - \csc{x}\cot{x}}{\csc{x}+\cot{x}} \, dx =$

$\ln|\csc{x}+\cot{x}| + C$

also, note that ...

$\csc{x} + \cot{x} = \frac{1+\cos{x}}{\sin{x}}$
Isn't $\ln | \csc x + \cot x | = - \ln | \csc x - \cot x |$ ?

11. ## Re: Simplification of -ln| cscx - cotx |

Originally Posted by Danny
Isn't $\ln | \csc x + \cot x | = - \ln | \csc x - \cot x |$ ?
you and the OP are correct, Danny ...

$-\ln| \csc{x} - \cot{x} | = -\ln \left|\frac{1-\cos{x}}{\sin{x}}\right| = \ln \left|\frac{\sin{x}}{1-\cos{x}}\right| = \ln\left|\frac{\sin{x}(1+\cos{x})}{1-\cos^2{x}}\right| = \ln\left|\frac{1+\cos{x}}{\sin{x}}\right|$

... my apologies.