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Question: If \(I_{n} = \int_{}^{}\frac{\sin nx}{\sin x}dx,\) where \(n > 1,\) then \(I_{n} - I_{n - 2} =\)...

If In=∫sin⁡nxsin⁡xdx,I_{n} = \int_{}^{}\frac{\sin nx}{\sin x}dx, where n>1,n > 1, then In−In−2=I_{n} - I_{n - 2} =

A

2(n−1)cos⁡(n−1)x\frac{2}{(n - 1)}\cos(n - 1)x

B

2n−1sin⁡(n−1)x\frac{2}{n - 1}\sin(n - 1)x

C

2ncos⁡nx\frac{2}{n}\cos nx

D

2nsin⁡nx\frac{2}{n}\sin nx

Answer

2n−1sin⁡(n−1)x\frac{2}{n - 1}\sin(n - 1)x

Explanation

Solution

In=∫sin⁡nxsin⁡xdxI_{n} = \int_{}^{}\frac{\sin nx}{\sin x}dx

In−2=∫sin⁡(n−2)xsin⁡xdx⇒In−In−2=∫sin⁡nx−sin⁡(n−2)xsin⁡xdxI_{n - 2} = \int_{}^{}{\frac{\sin(n - 2)x}{\sin x}dx} \Rightarrow I_{n} - I_{n - 2} = \int_{}^{}{\frac{\sin nx - \sin(n - 2)x}{\sin x}dx} =∫2cos⁡(n−1)x.sin⁡xsin⁡xdx= \int_{}^{}{\frac{2\cos(n - 1)x.\sin x}{\sin x}dx}

In−In−2=2sin⁡(n−1)x(n−1)I_{n} - I_{n - 2} = \frac{2\sin(n - 1)x}{(n - 1)}