If n is a positive integer greater than $\mathrm{I}$ and ${\mathrm{I}}_{\mathrm{n}}=\int \frac{\mathrm{sinnx}}{\mathrm{sinx}}\mathrm{dx}$, then ${\mathrm{I}}_{\mathrm{n}+1}-{\mathrm{I}}_{\mathrm{n}-1}=$

$I_n=\int \frac{\sin nx}{\sin x}dx\text{. }$

$$\begin{gathered} I_2=\int \frac{\sin 2x}{\sin x}dx=\int \frac{2\sin x\cos x}{\sin x}dx \\ =2\sin x+c \end{gathered}$$

$I_3=\int \frac{\sin 3x}{\sin x}dx=\int \frac{3\sin x-4{\sin }^{3}x}{\sin x}dx$

$\begin{array}{r}{I}_3=\int \left[3-4\left(\frac{1-\cos 2x}{2}\right)\right]dx\\ I_3=\int (1+2\cos 2x)dx=x+\frac{2}{2}\sin 2x+c\cdot \end{array}$

similarly

${\mathrm{I}}_{\mathrm{n}+1}-{\mathrm{I}}_{\mathrm{n}-1}=$$\frac{2}{\mathrm{n}}\mathrm{sinnx}$