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**The problem statement**

Find: [itex]\displaystyle\int \cot x \ \mathrm{d}x[/itex]

**A Solution**

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

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

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

We convert our result into a formula with respect to [itex]x[/itex]:

[itex]\boxed{\ln |\sin x| + \mathcal{C}}[/itex]

**My Question**

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

[itex]\displaystyle\int u \ \mathrm{d}v = uv - \displaystyle\int v \ \mathrm{d}u,[/itex]

we get [itex]1=0[/itex]. I can show my solution, if needed. Why is this? (A friend of mine pointed this out to me)Thanks in advance! :)