IMC 1996 · Day 1 · P2

We have $$\begin{align*} I_n &= \int_{-\pi}^{\pi} \frac{\sin nx}{(1 + 2^x) \sin x}\,dx \\ &= \int_0^{\pi} \frac{\sin nx}{(1 + 2^x) \sin x}\,dx + \int_{-\pi}^{0} \frac{\sin nx}{(1 + 2^x) \sin x}\,dx. \end{align*}$$ In the second integral we make the change of variable $x = -x$ and obtain $$\begin{align*} I_n &= \int_0^{\pi} \frac{\sin nx}{(1 + 2^x) \sin x}\,dx + \int_0^{\pi} \frac{\sin nx}{(1 + 2^{-x}) \sin x}\,dx \\ &= \int_0^{\pi} \frac{(1 + 2^x) \sin nx}{(1 + 2^x) \sin x}\,dx \\ &= \int_0^{\pi} \frac{\sin nx}{\sin x}\,dx. \end{align*}$$ For $n \ge 2$ we have $$\begin{align*} I_n - I_{n-2} &= \int_0^{\pi} \frac{\sin nx - \sin{(n-2)x}}{\sin x}\,dx \\ &= 2 \int_0^{\pi} \cos{(n-1)x}\,dx = 0. \end{align*}$$ The answer $$I_n = \begin{cases} 0 & \text{if $n$ is even,} \\ \pi & \text{if $n$ is odd} \end{cases}$$ follows from the above formula and $I_0 = 0$, $I_1 = \pi$.