Prove recurrent formula $(n+2)\int_{0}^{\pi} \ln{(\sin{\frac{x}{2}})}\cos{(n+2)x}dx = n\int_{0}^{\pi} \ln{(\sin{\frac{x}{2}})}\cos{nx}dx$ As stated in the title, I want to prove the recurrent formula $$(n+2)\int_{0}^{\pi} (\ln{(\sin{\frac{x}{2}})})\cos{(n+2)x}dx = n\int_{0}^{\pi} (\ln{(\sin{\frac{x}{2})})}\cos{nx}dx$$
for $n \in \mathbb{N}, n \geq 1$.
So far, I've tried partial integration, and I've managed to reduce the equality to $$\int_{0}^{\pi} \cot{\frac{x}{2}}\sin{(n+2)x}dx = \int_{0}^{\pi} \cot{\frac{x}{2}}\sin{nx}dx.$$
However, I'm not sure how to proceed from here.
Also, wolframalpha prints out $-\frac{\pi}{2}$ for any $n$ I plug in, which tells me that the equality almost definitely holds, but I can't figure out how to prove it.
 A: I had success when I shifted strategies, to prove that $\int_{0}^{\pi} \cot{\frac{x}{2}}\sin{nx}dx$ has a constant value - this was much easier. We use the identity $\sin(\alpha) + \sin(\beta) = 2\sin(\frac{\alpha + \beta}{2})\cos(\frac{\alpha - \beta}{2})$, and proceed by induction: \begin{eqnarray} \int_{0}^{\pi} (\cot{\frac{x}{2}}\sin{(n+1)x} - \cot{\frac{x}{2}}\sin{nx} )dx &=& 2\int_{0}^{\pi}\cot(\frac{x}{2})\sin(\frac{x}{2})\cos((n+\frac{1}{2})x)dx\\ &=& 2\int_{0}^{\pi}\cos(\frac{x}{2})\cos((n+\frac{1}{2})x)dx.\end{eqnarray}
Now we use the "evenness" of cosine to obtain $$\int_{-\pi}^\pi \cos(\frac{x}{2})\cos((n+\frac{1}{2})x)dx$$
and finally we use the identity $\cos(\alpha)\cos(\beta) = 
\frac{1}{2}(\cos(\alpha + \beta) + \cos(\alpha - \beta))$ to obtain
$$\int_{-\pi}^\pi \cos(n+1)x + \cos nxdx$$
and this integral is clearly $0$. Thus, each two successive terms of your integral (i.e., for successive values of $n$) have difference $0$, and so it follow by induction that they are all equal, for any $n$. 
