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.
Best Answer
Note
$$2\sin\frac x2(\cos x + \cos2x+\cos3x+...+\cos10x) = \sin\frac{21x}2-\sin\frac x2 $$
Then,
$$\begin{align} \int_0^\pi \frac{\sin\frac{21x}{2}}{\sin \frac x2}{\rm d}x =&\int_0^\pi(1+2\cos x + 2\cos2x+...+2\cos10x){\rm d}x\\ =&\pi + (0+0+...+0)\\ =&\pi \end{align}$$