Question

If $U_n={\int }_0^{\pi }\frac{1-\cos nx}{1-\cos x}dx$, where $n$ is a non-negative integer, then which of the following is/are true

• A. $U_{n+2}+U_n-2U_{n+1}=0$
• B. $U_0+U_1=2\pi$
• C. $U_n=n\pi$
• D. $U_{n+2}+U_n-2U_{n+1}=\pi$

Answer 1

Answer: (A), (C)

$U_n=n\pi$

$U_{n+2}-U_{n+1}={\int }_0^{\pi }\frac{(1-\cos (n+2\left)x\right)-(1-\cos (n+1\left)x\right)}{1-\cos x}dx={\int }_0^{\pi }\frac{\cos (n+1)x-\cos (n+2)x}{1-\cos x}={\int }_0^{\pi }\frac{2\sin (n+\frac{3}{2})x\sin \frac{x}{2}}{2{\sin }^2\frac{x}{2}}dx\Rightarrow U_{n+2}-U_{n+1}={\int }_0^{\pi }\frac{\sin (n+\frac{3}{2})x}{\sin \frac{x}{2}}dx\cdots \left(i\right)\Rightarrow U_{n+1}-U_n={\int }_0^{\pi }\frac{\sin (n+\frac{1}{2})x}{\sin \frac{x}{2}}dx\cdots \left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$, we get
$(U_{n+2}-U_{n+1})-(U_{n+1}-U_n)={\int }_0^{\pi }\frac{\sin (n+\frac{3}{2})x-\sin (n+\frac{1}{2})x}{\sin \frac{x}{2}}dx\Rightarrow U_{n+2}+U_n-2U_{n+1}={\int }_0^{\pi }\frac{2\cos (n+1)x\sin \frac{x}{2}}{\sin \frac{x}{2}}dx\Rightarrow 2{\int }_0^{\pi }\cos (n+1)xdx=2{\left(\frac{\sin (n+1)x}{n+1}\right)}_0^{\pi }=0\Rightarrow U_{n+2}+U_n=2U_{n+1}{U}_0={\int }_0^{\pi }\frac{1-1}{1-\cos x}dx=0,U_1={\int }_0^{\pi }\frac{1-\cos x}{1-\cos x}dx=\pi U_1-U_0=\pi \Rightarrow U_n=U_0+n\pi =n\pi \Rightarrow U_n=n\pi$