Prove that $\int_0^u\frac{\sin(mt)}{2\sin(t/2)}dt$ is bounded When I read the Fourier Series theory, a property show in the book but without the details.
That property is:

$\displaystyle\int_0^u\frac{\sin(mt)}{2\sin(t/2)}dt$ is bounded for
  all $m$ and $0\leq u\leq\pi$

I want to ask how to prove it? Also, what is the situation we need this theorem? Maybe it can simply be found on other books on this topic, but I don't have them.

Notice that, in general, for $n \ge 0$
$$\cos nx + \Bbb i \sin nx = \Bbb e ^{\Bbb i n x} = (\Bbb e ^{\Bbb i x})^n = (\cos x + \Bbb i \sin x)^n = \sum _{k=0} ^n \binom n k \Bbb i ^k \cos ^{n-k} x \sin ^k x$$
whence equating the imaginary parts in both sides gives us
$$\sin nx = \binom n 1 \cos ^{n-1} x \sin x - \binom n 3 \cos ^{n-3} x \sin ^3 x + \binom n 5 \cos ^{n-5} x \sin ^5 x - \\ \binom n 7 \cos ^{n-7} x \sin ^7 x + \dots = (\sin x) P_{n-1} (\cos x)$$
with $P_{n-1} = a_{n-1,n-1} x^{n-1} + a_{n-1,n-2} x^{n-2} + \dots + a_{n-1,0}$ a polynomial function of degree $n-1$.
It follows that
$$\sin mt = \sin \left( 2m \frac t 2 \right) = \left( \sin \frac t 2 \right) P_{2m-1} \left( \cos \frac t 2 \right) ,$$
whence, using the obvious fact that $\int _a ^b f  \le \int _a ^b |f|$,
$$\left| \int \limits _0 ^u \frac {\sin mt} {\sin \frac t 2} \ \Bbb d t \right| = \left| \int \limits _0 ^u \frac {\left( \sin \frac t 2 \right) P_{2m-1} \left( \cos \frac t 2 \right)} {\sin \frac t 2} \ \Bbb d t \right| = \left| \int \limits _0 ^u P_{2m-1} \left( \cos \frac t 2 \right) \ \Bbb d t \right| \le \int \limits _0 ^u \left| P_{2m-1} \left( \cos \frac t 2 \right) \right| \ \Bbb d t \le \\
\sum _{k=0} ^{2m-1} |a_{2m-1,k}| \int \limits _0 ^u \left| \cos ^k \frac t 2 \right| \ \Bbb d t \le \sum _{k=0} ^{2m-1} |a_{2m-1,k}| \int \limits _0 ^u 1 \ \Bbb d t = \sum _{k=0} ^{2m-1} |a_{2m-1,k}| u \le \sum _{k=0} ^{2m-1} |a_{2m-1,k}| \pi$$
which is a constant independent of $u$.

Of course, the much simpler approach would be to notice that the only point where the integrand might raise problems is $t=0$, but there
$$\lim _{t \to 0} \frac {\sin mt} {\sin \frac t 2} = 2m ,$$
so using the comparison test for improper integrals your integral must converge, i.e. be finite for all $0 \le u \le \pi$. Since the result of the integration is continuous with respect to $u$, and since $[0, \pi]$ is compact, Weiertrass's extreme value theorem tells us that your integral is bounded and attains its bounds.
