# sir please tell what is the integration of sin9x / sinx?

$\hspace{-0.6 cm}\int\frac{\sin 9x}{\sin x}dx = \int\frac{\sin 9x-\sin 7x+\sin 7x-\sin 5x+\sin 5x-\sin 3x+\sin 3x-\sin x+\sin x}{\sin x}\\\\\\Using \sin C-\sin D=2\cos \frac{C+D}{2}\sin \frac{C-D}{2}\\\\\\So \int\frac{\sin 9x}{\sin x}dx = \int \left(\cos 8x+\cos 6x+\cos 4x+\cos 2x+1\right)dx\\\\\\So \int\frac{\sin 9x}{\sin x}dx = \frac{\sin 8x}{8}+\frac{\sin 6x}{6}+\frac{\sin 4x}{4}+\frac{\sin 2x}{2}+x+\mathcal{C}$