# Thread: integration question no. 7

1. ## integration question no. 7

integration question no. 7

2. $\displaystyle I_n=\int_0^{\pi}\frac{\sin (2n-1)x}{\sin x}dx=\int_0^{\pi}\frac{\sin 2nx\cos x-\sin x\cos 2nx}{\sin x}dx=$
$\displaystyle=\int_0^{\pi}\frac{\sin 2nx\cos x}{\sin x}dx-\int_0^{\pi}\cos 2nxdx=$
$\displaystyle=\frac{1}{2}\int_0^{\pi}\frac{\sin(2n +1)x+\sin(2n-1)x}{\sin x}dx-\left.\frac{1}{2n}\sin 2nx\right|_0^{\pi}=$
$\displaystyle=\frac{1}{2}I_{n+1}+\frac{1}{2}I_n\Ri ghtarrow\frac{1}{2}I_n=\frac{1}{2}I_{n+1}\Rightarr ow I_n=I_{n+1}$

3. Hello, afeasfaerw23231233!

Let: . $I_n \;=\;\int^{\pi}_0\frac{\sin(2n+1)x}{\sin x}\,dx$

Bu using the identity: . $\sin(A - B) \:=\:\sin A\cos B - \sin B\cos A$

. . show that: . $I_{n+1} \:=\:I_n$
I will drop the limits (for now) . . .

We have: . $I_n \;=\;\int \frac{\sin(2nx - x)}{\sin x}\,dx \;=\;\int\frac{\sin(2nx)\cos(x) - \sin(x)\cos(2nx)}{\sin x}\,dx$

. . $I_n \;=\;\int\left[\frac{\sin(2nx)\cos(x)}{\sin(x)} - \cos(2nx)\right]\,dx \;=\;\int\frac{\sin(2nx)\cos(x)}{\sin(x)}\,dx \,- \int \cos(2nx)\,dx$

We have: . $I_{n+1} \;=\;\int\frac{\sin(2nx + x)}{\sin x}\,dx \;=\;\int\frac{\sin(2nx)\cos(x) + \sin(x)\cos(2nx)}{\sin x}\,dx$

. . $I_{n+1} \;=\;\int\left[\frac{\sin(2nx)\cos(x)}{\sin x} + \cos(2nx)\right]\,dx \;=$ . $\int\frac{\sin(2nx)\cos(x)}{\sin x}\,dx + \int \cos(2nx)\,dx$

We see that $I_n$ and $I_{n+1}$ are identical except for their second integrals.

. . But: . $\int^{\pi}_0\cos(2nx)\,dx \;=\;\frac{1}{2n}\sin(2nx)\,\bigg|^{\pi}_0\;=\;\fr ac{1}{2n}\left[\sin(2n\pi) - \sin(0)\right] \;=\;0$

Therefore: . $I_{n+1} \;=\;\int^{\pi}_0\frac{\sin(2xn)\cos(x)}{\sin x}\,dx \;=\;I_n$