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This problem came from some other website, where someone asked for help with the integral

$$\int\frac {x \, dx}{1-\cos x}$$

After adding my suggestion of integration by parts to an existing suggestion of multiplying top and bottom by $1+\cos x$, he thanked us for his help and posted his solution. Due to a trig identity I didn't recognize and formatting issues with the lack of latex, I posted my own solution. I see his logic now, but what I don't see are how the 2 results are equivalent.

Both methods start the same way by multiplying top and bottom by $1+\cos x$.

$$\int\frac{x(1+\cos x) \, dx}{\sin^2x \, dx}=\int x(\csc^2x+\csc x\cot x) \, dx$$

My method was to use integration by parts here.

$$u=x,\quad du=dx$$ $$dv=(\csc^2x+\csc x\cot x) \, dx,\quad v=-\cot x -\csc x$$ $$\int x(\csc^2x+\csc x\cot x) \, dx=-x\cot x-x\csc x+\int(\cot x +\csc x) \, dx=$$ $$-x\cot x-x\csc x-\int\frac{-\cos x}{\sin x} \, dx-\int\frac{-\cot x\csc x-\csc^2x}{\cot x+\csc x} \, dx=$$ $$-x\cot x-x\csc x-\ln(\sin x)-\ln(\cot x+\csc x)+C=$$ $$-x\cot x-x\csc x+\ln(\csc x)+\ln\frac1{\cot x+\csc x}+C=$$ $$-x\cot x-x\csc x+\ln(\csc x)+\ln\frac{\csc x-\cot x}{\csc^2x-\cot^2x}+C=$$ $$-x\cot x-x\csc x+\ln(\csc^2x-\csc x\cot x)+C$$

Hopefully, the next method will be shorter. He also did integration by parts with the same substitutions, but used the reciprocal of a tangent half-angle formula to obtain

$$v=-\csc x-\cot x=-\cot\frac x2$$

making the result of his integration by parts

$$-x\cot\frac x2+\int\cot\frac x2=-x\cot\frac x2-2\int\frac{-\frac12\cos\frac x2}{\sin\frac x2}=$$

$$-x\cot\frac x2-2\ln\left(\sin\frac x2\right)$$

Given the identity he used, the first part seems to agree, but I can't seem to show that

$$-2\ln\left(\sin\frac x2\right)=\ln(\csc^2x-\csc x\cot x)+C$$

I tried the following

$$-2\ln\left(\sin\frac x2\right)=\ln\left(\frac1{\sin^2\frac x2}\right)=\ln\left(\frac2{1-\cos x}\right)=\ln\left(\frac{2+2\cos x}{\sin^2x}\right)=\ln(2\csc^2x+2\csc x\cot x)$$

which does not agree. So where is the mistake?

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    $\begingroup$ The integral of cotangent is ln|sin(x)| $\endgroup$ – The Chaz 2.0 Oct 18 '12 at 3:38
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(For the sake of having an answer...)

Your mistake is when you integrated the cotangent. I don't know why you "took out" a negative sign, but it should just be $\ln | \sin x|$

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  • $\begingroup$ oops! Normally, the cofunctions are the ones with the minus signs. I'll have to go back through it fixed later, but that's probably the problem. $\endgroup$ – Mike Oct 18 '12 at 5:32
  • $\begingroup$ That was the problem. He actually had the correct answer before I "corrected" it here. After correcting the cotangent integration, both answers agreed. $\endgroup$ – Mike Oct 18 '12 at 22:40
  • $\begingroup$ Good deal. $$$$ $\endgroup$ – The Chaz 2.0 Oct 18 '12 at 22:54

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