Prove the following identity where $n\in\mathbb{N}$ $$\int\frac{dx}{\sin^{n-1}(x)\prod_{i=1}^n(\cot(x)-\cot((i+1)x))}=\frac{-\cos((n+1)x)}{n+1}$$

This was a subpart in a massive problem. I was able to solve everything except this. According to wolframalpha I guess that this result is true but I am not able to prove it. Can anyone help me cause this small thing is preventing me from proceeding further in the massive problem.

Any help is greatly appreciated.


1 Answer 1


First, by the difference of cotangents formula, we have for an arbitrary $i$ that $$\begin{align*} \cot(x) - \cot([i+1]x) &= \cot(x) - \frac{\cot(ix)\cot(x) - 1}{\cot(ix) + \cot(x)}\\ &= \frac{\cot^2(x) + 1}{\cot(ix) + \cot(x)} \\ &= \frac{\csc^2(x)}{\cot(ix) + \cot(x)} \end{align*}$$ Then, the denominator of the integral is $$\begin{align*} \sin^{n-1}(x)\prod_{i=1}^n \left\{\cot(x) - \cot([i+1]x)\right\} &= \frac{1}{\sin^{n+1}(x)}\prod_{i=1}^n\frac{1}{\cot(ix) + \cot(x)} \\ &= \frac{1}{\sin^{n+1}(x)}\prod_{i=1}^n\frac{\sin(x)\sin(ix)}{\sin(ix)\cos(x) + \cos(ix)\sin(x)} \\ &= \frac{1}{\sin(x)}\prod_{i=1}^n\frac{\sin(ix)}{\sin([i+1]x)} \\ &= \frac{1}{\sin(x)}\frac{\sin(x)}{\sin([n+1]x)} \end{align*}$$ And then the integral is $$\begin{align*} \int \frac{\text{d}x}{\sin^{n-1}(x)\prod_{i=1}^n \left\{\cot(x) - \cot([i+1]x)\right\}} &= \int \sin([n+1]x)\text{d}x \\ &= - \frac{\cos([n+1]x)}{n+1}. \end{align*}$$


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