Evaluate $\int_0^\pi \frac{\sin\frac{21x}{2}}{\sin \frac x2} dx$ (from MIT Integration Bee) 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.
 A: Note that if we call $a=e^{ix/2}$, then we have $$\frac{\sin(21x/2)}{\sin(x/2)} = \frac{a^{21}-a^{-21}}{a-a^{-1}} = a^{-20}\frac{a^{42}-1}{a^2-1}=a^{-20}\frac{(a^2-1)(a^{40}+a^{38}+a^{36}+\cdots+1)}{a^2-1}=(a^{20}+a^{18}+\cdots+a^{-18}+a^{-20})$$
Then since $a^n+a^{-n}=2\cos(nx/2)$ and $\int_0^\pi \cos(nx/2) dx=0$ for $n$ even, all but the $a^0$ term of the above product vanishes under integration.  Hence the integral is just $\int_0^\pi 1 dx=\pi$
A: 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.
A: I found another elementary solution :
We have, as from the post, $$I= \int_0^\pi \frac{\sin 11x}{\sin x} dx$$
Making the substitution $x \mapsto \frac x2 \implies dx \mapsto \tfrac 12 dx$
$$I=\frac 12 \int_0^{2\pi} \frac{\sin \frac{11x}{2}}{\sin \frac x2} dx$$ Multiplying top and bottom by $\cos \frac x2$, 
$$I = \frac 12 \int_0^{2\pi} \frac{\sin 6x + \sin 5x}{\sin x} dx$$
Using the fact that $\int_0^{2a} f(x)dx= 2\int_0^a f(x)dx$ if $f(x) = f(2a-x)$,
$$I= \int_0^\pi \frac{\sin 6x + \sin 5x}{\sin x}dx$$
$$ = \int_0^\pi \frac{\sin 5x}{\sin x}dx$$
Repeat these steps one more time to get
$$I= \int_0^\pi \frac{ \sin 3x}{\sin x}dx$$
$$= \int_0^\pi (3 - 4\sin^2x)\,dx = \pi$$
A: Note
$$2\sin\frac x2(\cos x + \cos2x+\cos3x+...+\cos10x)
= \sin\frac{21x}2-\sin\frac x2 $$
Then,
$$\begin{align}
\int_0^\pi \frac{\sin\frac{21x}{2}}{\sin \frac x2}{\rm d}x 
=&\int_0^\pi(1+2\cos x + 2\cos2x+...+2\cos10x){\rm d}x\\
=&\pi + (0+0+...+0)\\
=&\pi \end{align}$$
