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I manually integrate $\int \frac{1}{\sin(x)}dx$ as $$\int \frac{\sin(x)}{\sin^{2}(x)}dx = -\int \frac{1}{\sin^{2}(x)}d\cos(x) = \int \frac{1}{\cos^{2}(x) - 1}d\cos(x).$$

After replacing $u = \cos(x)$,

$$\int \frac{1}{u^{2} - 1}du = \int \frac{1}{u^{2} - 1}du = \frac {1} {2} \int \left(\frac {1} {u - 1} - \frac {1} {u+1}\right) du = \frac {1} {2} \ln\left(\frac {u-1} {u+1}\right) + C.$$

Substitute back to obtain

$$\frac {1} {2} \ln\left(\frac {\cos(x)-1}{\cos(x)+1}\right) + C.$$

The problem is that this solution is incorrect (I guess) because for example http://www.integral-calculator.com/ gives another solution

$$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$

And all other online solvers gives equivalent solution to $$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$

The question is there I made a mistake?

Update: some of you may say that in complex space my answer is right but not so fast:

Take wolfram solver: integrate 1/sinx

The we get: $-ln(cot(x) + csc(x)) + C$ It is easy to see that it is equvalent to $$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$

$-\ln(\cot(x) + \csc(x)) + C = -\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)})$

then

$-\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))^{2}} {\sin^{2}x}) = -\frac {1}{2} \ln(\frac{1+\cos(x)+\cos(x)+\cos^{2}(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))(\cos(x)+\cos^{2})(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {1+\cos(x)} {1-\cos(x)}) = \frac {1}{2} \ln(\frac {1-\cos(x)} {1+\cos(x)})$

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3 Answers 3

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Where did I make a mistake?

First, you rather have $$ \frac {1} {2} \int \left(\frac {1} {u - 1} - \frac {1} {u+1}\right) du = \frac {1} {2} \ln\left|\frac {u-1} {u+1}\right| + C $$then observe that $$ \left|\frac {\cos(x)-1}{\cos(x)+1}\right|=\frac {1-\cos(x)}{1+\cos(x)} $$ since $$ 1 - \cos(x)\ge 0,\quad 1+ \cos(x)\ge0, $$ giving that $$ \ln\left|\frac {\cos(x)-1}{\cos(x)+1}\right|=\ln\left(\frac {1-\cos(x)}{1+\cos(x)}\right). $$

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  • $\begingroup$ Ok It is a point but that if I calculate this in complex space? Then $\int \frac {1}{x} dx = ln(x)$ not $ln|x|$ (I guess) $\endgroup$ Mar 11, 2017 at 10:16
  • $\begingroup$ @MaratZakirov Here we assume to integrate in a real space. In a complex space, you are right to say it does not hold. $\endgroup$ Mar 11, 2017 at 10:17
  • $\begingroup$ @MaratZakirov How do you define $\log(z)$ for $z \in \mathbb{C}$? Thanks. One may recall that no branch of $\log(z)$ exists on an open set $U$ containing a closed curve that winds around $0$... $\endgroup$ Mar 11, 2017 at 10:26
  • $\begingroup$ please see update of my question in complex space my answer is still incorrect and pay attention to fact that it is done on wolfram alpha solver which is complex by default $\endgroup$ Mar 11, 2017 at 10:38
  • $\begingroup$ @MaratZakirov I must confess that I don't really understand your point here ;) $\endgroup$ Mar 11, 2017 at 10:40
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If you are considering antiderivatives as functions on connected domains $D$ that are open in $\mathbb{C}$ -- suitably chosen so that we don't have to worry about multivaluedness of the logarithm -- then you should still be okay. It's still true that the functions $z \mapsto \frac1{2} \ln \left(\frac{1 - \cos(z)}{1 + \cos(z)}\right)$ and $z \mapsto \frac1{2} \ln \left(\frac{\cos(z) - 1}{\cos(z) + 1}\right)$ are antiderivatives of $z \mapsto 1/\sin(z)$. Their difference is the constant $C = \ln(-1)/2$ for whatever branch of the logarithm you are considering. So you still get the same family of functions, up to a complex constant.

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  • $\begingroup$ EXACTLY!!!!!!!!!!!!!!!!!!!!!! $\endgroup$ Mar 11, 2017 at 13:34
  • $\begingroup$ but ln(-1) is only suitable for complex space in real we should take module as previous answerers were mentioned $\endgroup$ Mar 11, 2017 at 13:46
  • $\begingroup$ Yes, that's indeed true. $\endgroup$
    – user43208
    Mar 11, 2017 at 14:39
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The integral of $1/x$ is not $\ln(x)$, but $\ln|x|$. See What is the integral of 1/x?

Then, in your case $\frac{1}{2}\int du \left( \frac{1}{u-1} - \frac{1}{u+1} \right) = \frac{1}{2}\ln\left( \frac{|u-1|}{|u+1|} \right) + C$. When you put back $u=\cos(x)$, the expression $|u-1|$ becomes $|\cos(x)-1|=1-\cos(x)$ and $|u+1|$ becomes $|\cos(x)+1|=1+\cos(x)$. Notice, you can forget about taking the absolute value, because $1-\cos(x) \ge 0$ and $1+\cos(x)\ge 0$. Finally

$\int dx \frac{1}{\sin(x)} = \frac{1}{2}\ln\left( \frac{1-\cos(x)}{1+\cos(x)} \right) + C$

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  • $\begingroup$ In real space you are right but what if you consider complex space? $\endgroup$ Mar 11, 2017 at 11:21

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