Reduction Formulae (High School) I've been going through working out some reduction formulae question and I've been finding this question particularly difficult.
Problem: 
Given:
$$I_n=\int \frac{\sin(2nx)}{\sin(x)} \, dx $$ to show that: $$I_n - I_{n-1}=\frac{2(-1)^{n+1}}{2n-1}\,$$
 A: Applying the Prosthapharesis formula
$$ 2\cos{a}\sin{b} = \sin{(a+b)}-\sin{(a-b)}, $$
(I find it easier to remember it this way round) with $a=(2n-1)x$ and $b=x$, so
$$ \sin{2nx}-\sin{(2nx-x)} = 2\cos{(2n-1)x}\sin{x}, $$
the integral becomes
$$ I_n-I_{n-1} = \int \frac{2\cos{(2n-1)x}\sin{x}}{\sin{x}} \, dx = \frac{2}{2n-1}\sin{(2n-1)x},  $$
at which point I suspect you probably mean $\int_0^{\pi/2}$, which gives
$$ \frac{2}{2n-1}\sin{(n-\tfrac{1}{2})\pi} -0 = \frac{2(-1)^{n+1}}{2n-1} $$
on the right.
A: Notice, given that $$I_n=\int \frac{\sin(2nx)}{\sin x}dx$$
setting $n=n-1$, we get 
$$I_{n-1}=\int \frac{\sin(2(n-1)x)}{\sin x}dx$$
$$\implies I_n-I_{n-1}=\int \frac{\sin(2nx)}{\sin x}dx-\int \frac{\sin(2(n-1)x)}{\sin x}dx$$
$$=\int \frac{\sin(2nx)-\sin(2(n-1)x)}{\sin x}dx$$
$$=\int \frac{2\cos\left(\frac{2nx+2(n-1)x}{2}\right)\sin\left(\frac{2nx-2(n-1)x}{2}\right)}{\sin x}dx$$
$$=\int \frac{2\cos\left(2nx-x\right)\sin x}{\sin x}dx$$ $$=2\int\cos (2n-1)xdx $$
Applying limits $x=0$ to $x=\frac{\pi}{2}$, we get 
$$I_n-I_{n-1}=2\int_{0}^{\pi/2}\cos (2n-1)xdx $$ let $(2n-1)x=t\implies dx=\frac{dt}{2n-1}$ $$=2\int_{0}^{(2n-1)\frac{\pi}{2}}(\cos t) \frac{dt}{(2n-1)} $$
$$= \frac{2}{2n-1}\int_{0}^{(2n-1)\frac{\pi}{2}}\cos t\ dt $$
$$= \frac{2}{2n-1}[\sin t]_{0}^{(2n-1)\frac{\pi}{2}}$$ $$=\frac{2}{2n-1}\sin\left((2n-1)\frac{\pi}{2}\right)$$
$$=\color{red}{\frac{2(-1)^{n+1}}{2n-1}}$$
