The teacher made this approach to solve the Dirichlet integral , $$ J_n= \int_0^\frac{\pi}{2} \frac{\sin(2nx)}{\sin x}\:\mathrm{d}x,\quad I_n = \int_0^\frac{\pi}{2} \frac{\sin(2n+1)x}{\sin x}\:\mathrm{d}x $$ - Calculate $$ J_n -J_{n-1}$$ and $$ I_n -I_{n-1} $$ -Conclude $$I_n , J_n$$
This exercises was part of the Dirichlet integral exercise, I did calculated $I_n $, $ I_n = \frac{\pi}{2}$. The problem I had with $J_n$; I used all the Trig Formulae I can think of, didn't get to simpler result like I did with $I_n$ where you find that $I_n -I_{n-1} = 0$ then easily conclude $I_n$.
I applied the formula mentioned above , Here's what I got : $$ J_n -J_{n-1} = \int_0^\frac{\pi}{2} \frac{2}{2n-1}cos(2n-1)xdx $$ So $$ J_n -J_{n-1} = \frac{2}{2n-1}sin[(2n-1) \frac{\pi}{2}] $$ We can remarque that 2n-1 is not pair that means $$ sin[(2n-1) \frac{\pi}{2}] =+ 1$$ or $$ -1 $$ we can write $$ J_n -J_{n-1}= \frac{2(-1)^n}{2n-1} $$ Then What should I do to find $$ J_n$$ ?