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?

  • 2
    $\begingroup$ The integral of cotangent is ln|sin(x)| $\endgroup$ – The Chaz 2.0 Oct 18 '12 at 3:38

(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 $\ln | \sin x|$

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.