IBP Paradox for cot x integral

[ahaanomegas]

Greetings! I hope this is the correct forum for my question (this is my first post, here).

The problem statement

Find: \displaystyle\int \cot x \ \mathrm{d}x

A Solution
We can re-write this integral into a convenient form: \displaystyle\int \cot x \ \mathrm{d}x = \displaystyle\int \dfrac{\cos x}{\sin x} \ \mathrm{d}x

Now, let u=\sin x. Then, du = \cos x \ \mathrm{d}x.

The integral becomes: \displaystyle\int \dfrac {du}{u} \ \mathrm{d}x and we integrate to get \ln |u| (ignoring the constant of integration, for now -- reason: simplicity).

We convert our result into a formula with respect to x:

\boxed{\ln |\sin x| + \mathcal{C}}​

My Question

If we use Integration by Parts, however, with u=\dfrac{1}{\sin x} and \mathrm{d}v = \cos x \ \mathrm{d}x and

\displaystyle\int u \ \mathrm{d}v = uv - \displaystyle\int v \ \mathrm{d}u,​

we get 1=0. I can show my solution, if needed. Why is this? (A friend of mine pointed this out to me)

Thanks in advance! :)

 

[dynamicsolo]

You get 1 = 0 only by rather slippery neglect of the arbitrary constant. The full integration-by-parts would have to read

\int \frac{cos x dx}{sin x} = ( csc x )( sin x ) + \int (sin x) ( csc x cot x ) dx \mathbf{+ C } .

Someone had a post a month or so back (I don't know if I can find it again quickly) where the paradox there also turned on making a "1" appear out of nowhere. What you learn about indefinite integration is that finite numbers don't count for much...

That sounds facetious, but you do find that, for example, certain trigonometric-powers integrals can give one form of an anti-derivative using one technique of integration, and a different (but equivalent) form of the anti-derivative using another method. When you apply trig identities to show the equivalence of the two results, you find that they differ by a small number. But since both results are general anti-derivatives with arbitrary constants, the small number is actually irrelevant (in a sense, it shows the difference in "vertical shift" of the two anti-derivative functions). So the specified constant value can be made to appear or disappear, depending on the choice of anti-derivative. One could say that arbitrary constants of integration "swallow" numerical constants.