TITLE: Evaluate $\int_0^\pi \frac{\sin\frac{21x}{2}}{\sin \frac x2} dx$ (from MIT Integration Bee)
QUESTION [12 upvotes]: I recently watched the MIT Integration Bee ($2006$) video and stumbled upon this unusual integral: $$\int_0^\pi \frac{\sin\frac{21x}{2}}{\sin \frac x2} dx$$
I thought multiplying up and down by $\cos \frac x2$ would help, after which I got 
$$ \int_0^\pi \frac{\sin11x + \sin10x}{\sin x}dx = I$$
Now using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,
$$I=\int_0^\pi \frac{\sin 11x -\sin 10x}{\sin x}$$ and on adding the two we get
$$I= \int_0^\pi \frac{\sin 11x}{\sin x}$$
Now there are two paths I could take, either write $\sin 11x$ entirely in terms of $\sin x$ (which is a daunting task) or apply the sine addition rule as $\sin 11x = \sin(10x + x)$. Doing the latter gives
$$I= \int_0^\pi \frac{\sin 10x}{\sin x} \cos x \space dx + \int_0^\pi \cos 10xdx$$
$$= \int_0^\pi \frac {\sin 10x}{\sin x} \cos x\space dx$$
Do I keep going from here by using the sine addition rule again? Or is there a better way? There probably is.

REPLY [5 votes]: Define 
$$I_m= \int_{0}^{\pi} \frac{\sin{(\frac{mx}{2})}}{\sin{(\frac{x}{2})}} dx $$
for each $m\in\mathbb{N}$.
Clearly, we have $I_1=\pi$. Now, for $m\in\mathbb{N}$, we have
\begin{align} I_{m+2}-I_m = \int_{0}^{\pi} \frac{\sin{(\frac{mx+2x}{2})}-\sin{(\frac{mx}{2})}}{\sin{(\frac{x}{2})}} dx &= \int_{0}^{\pi} 2\cos{\left(\frac{(m+1)x}{2}\right)} dx \\ &= \left(\frac{4}{m+1}\right) \sin{\frac{(m+1)\pi}{2}}\end{align}
Letting $m=2k-1$ for $k\in\mathbb{N}$ and summing over $k$, it follows that $I_{2k-1}=I_1=\pi$ for all $k\in\mathbb{N}$. Now, letting $k=11$ solves the problem.