<s>[INST] Evaluate $ \int_0^\pi \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos (2x)}\right) \mathrm{d}x $ Evaluate the following definite integral: 
$$ \int_0^\pi  \left( \frac{2 + 2\cos (x) - \cos((k-1)x) - 2\cos (kx) - \cos((k+1)x)}{1-\cos(2x)}\right) \mathrm{d}x, $$ where $k \in \mathbb{N}_{>0}$.
 [/INST] Let 
$$I_k = \int_0^\pi  \left( \frac{2 + 2\cos x - \cos(k-1)x - 2\cos kx - \cos(k+1)x}{1 - \cos2x}\right) \mathrm dx.$$
Then, we have:


*

*$I_0 = 0$;

*$I_1 = \pi$.
For any  $ k \in \mathbb{N}^*$, we have:
$$ \begin{align} 
I_{k+1} - 2I_k + I_{k-1} 
&=  \int_0^\pi  \left(  \frac{\cos(-2+k)x - 2\cos kx + \cos (2 + k)x}{-1 + \cos 2x}\right) \mathrm dx \\
&= \int_0^\pi  \left(  \frac{2\cos kx(-1 + \cos^2x)}{-1 + \cos 2x}\right) \mathrm dx =0
\end{align}$$
Since $ I_{k+1} - I_k = I_k - I_{k-1} $, $(I_k)$ is an arithmetic progression. Hence $ I_k  = k\pi $.
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