It seems that $$\int_0^\pi \cot(x/2)\sin(nx)\,dx=\pi$$ for all positive integers $n$.
But I have trouble proving it. Anyone?
Answer
Use this famous sum:
$$1+2\cos x+2\cos 2x+\cdots+2\cos nx=\frac{\sin (n+\frac{1}{2})x}{\sin \frac{x}{2}}=\sin nx\cot\left(\frac{x}{2}\right)+\cos nx$$
Hence
$$\int_0^{\pi}\cot \left(\frac{x}{2}\right)\sin n x\,dx=\int_0^{\pi}1+2\cos x+2\cos 2x+\cdots +\cos nx\,dx$$
All cosine terms obviously evaluate to zero.
No comments:
Post a Comment