Enjoying 2017?

Calculus Level 4

$\large \int_{0}^{\pi}\frac{\sin(2017x)}{\sin x} \, dx = \, ?$

\frac{\pi}{2}

\pi

2\pi

1 solution

Kalpok Guha
Jul 24, 2017

$I_n=\int_{0}^{\pi}\frac{sin(2n+1)x}{sinx}$

Now $I_n-I_{(n-1)}= \int_{0}^{\pi}\frac{ sin(2n+1)x- sin(2n-1)x}{sinx}$

Or, $I_n-I_{(n-1)}= \int_{0}^{\pi}\frac{2cos2nxsinx}{sinx}$

Or, $I_n-I_{(n-1)}= 2\int_{0}^{\pi}\ cos2nx$

Or, $I_n-I_{(n-1)}=0$

Or, $I_n=I_{(n-1)}$

$I_0=I_1=...=I_{(2017)}=\pi$