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I was wondering if my solution to the integral: $$\int\frac{dx}{1 - \cos x}$$ is legit?

$$ \int\frac1{\cos x-1}\,\mathrm{d}x=\int-\frac12\cdot\frac1{\sin^2\left(\frac x2\right)}=\frac12\int-\csc\left(\frac12x\right)\,\mathrm{d}x=\boxed{\cot\left(\frac12x\right)+C}\\ \left(\cot\left(\frac12x\right)\right)'=-\csc^2\left(\frac12x\right)\cdot\frac12=\boxed{-\frac12\csc^2\left(\frac12x\right)}\\ \cos x-1=\cos x-\cos(0)=-2\sin\left(\frac{x+0}2\right)\sin\left(\frac{x-0}2\right)=-2\sin^2\left(\frac x2\right) $$ orginal image

My solution is based around the fact that the derivative of $\cot x$ is $-\csc^2x$. I basically converted $\cos x-1$ to $\cos x - \cos 0$ and from there used the $\cos a - \cos b$ trig identity.

I then googled the solution and found that there is a way to do it by multiplying by a conjugate but was still wondering if there is a flaw in my logic which I don't realize?

Thanks in advance :)

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    $\begingroup$ Please consider transcribing to MathJax when you find some time. $\endgroup$ Commented Jan 30, 2022 at 22:57
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    $\begingroup$ I am not sure why there are so many close votes for lack of context. The OP has supplied a valid method of integration and asked for verification. Their work is in an image, but there is nothing keeping anyone from improving the question by adding latex. $\endgroup$
    – robjohn
    Commented Jan 31, 2022 at 7:52
  • $\begingroup$ Thanks @robjohn. Next time I'll use MathJax to format my question. I'm new here and didn't know how it was called. $\endgroup$ Commented Jan 31, 2022 at 16:15

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Your logic is good, and your answer is correct. In the video, they find that $\int \frac{dx}{1-\cos x} = \csc x + \cot x + C$, and you found that $\int \frac{dx}{1-\cos x} = \cot \frac{x}{2} + C$, and you can show that these are equivalent:

$$\begin{eqnarray}\csc x + \cot x & = & \frac{1}{\sin x} + \frac{\cos x}{\sin x} \\ & = & \frac{1 + \cos x}{\sin x} \\ & = & \frac{1 + (2 \cos^2 \frac{x}{2} - 1)}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\ & = & \frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\ & = & \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \\ & = & \cot \frac{x}{2} \end{eqnarray}$$

(Note that there was no guarantee that the $+C$ in both integrals would be the same, but in this case it is.)

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  • $\begingroup$ Thank you. Nice way to verify the result :) $\endgroup$ Commented Jan 30, 2022 at 22:52

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