IBP Paradox for cot x integral

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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! :)
 
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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.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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