# IBP Paradox for cot x integral

• ahaanomegas

#### 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)

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