Find a reduction formula for $\large \rm ∫\frac{\sin(nx)}{\sin(x)} $

Question

faiqraees · Answer

Consider $\large \rm  I_{n+2}-I_{n} $

kainui · Answer

I was thinking, well ok maybe that means this has a nice closed form:

$$\int \frac{\sin((n+2)x)-\sin(nx)}{\sin x}dx$$
But I can't seem to find anything particularly illuminating hmm.

p0sitr0n · Answer

I think this might be the case when you can differentiate twice, then you get the same expression +- something on both sides, like I=sin(x)-7/n*I, then you send it to the other side?

faiqraees · Answer

$\color{#0cbb34}{\text{Originally Posted by}}$ @P0sitr0n
I think this might be the case when you can differentiate twice, then you get the same expression +- something on both sides, like I=sin(x)-7/n*I, then you send it to the other side?
$\color{#0cbb34}{\text{End of Quote}}$
Exactly. But I think we have to integrate I(n+2) and I(n) both

parthkohli · Answer

Most of the work is done.$$\frac{\sin(n+2)x - \sin nx}{\sin x} = \frac{2 \cos (n+1)x \sin x}{\sin x} = 2 \cos (n+1)x$$

faiqraees · Answer

$\color{#0cbb34}{\text{Originally Posted by}}$ @ParthKohli
Most of the work is done.$$\frac{\sin(n+2)x - \sin nx}{\sin x} = \frac{2 \cos (n+1)x \sin x}{\sin x} = 2 \cos (n+1)x$$
$\color{#0cbb34}{\text{End of Quote}}$
Can you tell me how you approached to the numerator in the second equality?

parthkohli · Answer

$$\sin C-\sin D = 2 \cos \frac{C+D}2\sin\frac{C-D}2 $$

faiqraees · Answer

Oh thanks