A tough integral and its generalization: I happened to encounter an integral, a definite while I was walking the other day:
$$ \int\limits_0^{\pi} \frac{ \sin ( 100 t ) }{\sin t } dt $$
I have tried the usual methods and nothing. I have tried to use an even-odd argument, and again with no exit. I know it is an improper integral, but I cannot see the trick: Also, can we generalize this integral to the cases: (n > 1) integer is n
$$ \int\limits_0^{\pi} \frac{ \sin (nt) }{\sin t } dt , \; \; \; \; \int\limits_0^{\pi} \frac{ \cos (nt) }{ \cos t } dt $$
 A: This question has been posted before, but here is a solution for the sine version:
$$ I_n = \int_0^\pi \frac{\sin(nt)}{\sin t} dt \\ I_{n+2}-I_n =\int_0^\pi \frac{\sin((n+2)t)-\sin(nt)}{\sin t} dt $$
Using $ \ \sin a - \sin b = 2 \sin(\frac{a-b}{2})\cos(\frac{a+b}{2})$:
$$I_{n+2}-I_n = 2 \int_0^\pi \frac{\sin t\cos((n+1)t)}{\sin t} dt = 2\int_0^\pi \cos((n+1)t)dt =0$$
Thus $\ I_{n+2}=I_n \ for \ n \ge 2 \ $ and it only suffices to evaluate the integral for n=2 and n=3 and all other larger values of n follow by induction.
$$ I_2 = \int_0^\pi \frac{\sin 2t}{\sin t} dt = 2\int_0^\pi \frac{\sin t \cos t}{\sin t} dt = 2\int_0^\pi \cos t dt = 0 \ $$ For I3, use the same identity as above but rearranged: $ \ \sin 3t - \sin t= 2\cos 2t \sin t  \ \therefore \ \sin 3t = \sin t (2 \cos 2t +1)$
$$ I_3 = \int_0^\pi \frac{\sin 3t}{\sin t} dt =\int_0^\pi 2 \cos 2t+1 dt  =\pi \\ I_n = 0 \  for  \ even \ n \ge 2 \\ I_n = \pi \  for  \ odd \ n \ge 3 \\ \therefore \int_0^\pi \frac{\sin 100t}{\sin t} dt = 0$$
A: Stupid idea: Use http://de.wikipedia.org/wiki/Formelsammlung_Trigonometrie#Winkelfunktionen_f.C3.BCr_weitere_Vielfache
Then you get
$$\int\limits_0^{\pi} \frac{ \sin (nt) }{\sin t } dt
=
\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor} (-1)^k {n-k-1 \choose k} 2^{n-2k-1} \int\limits_0^{\pi} \cos^{n-2k-1} (t)\ dt$$
What a nice "solution"!
