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(D) We are given $I_n = \int \frac{\sin nx}{\sin x} dx$. Consider the difference $I_n - I_{n-2}$: $I_n - I_{n-2} = \int \frac{\sin nx - \sin(n-2)x}{\s

Vedclass · Accepted Answer

$\frac{2}{n} \sin n x$

Vedclass · Answer

(D) We are given $I_n = \int \frac{\sin nx}{\sin x} dx$. Consider the difference $I_n - I_{n-2}$: $I_n - I_{n-2} = \int \frac{\sin nx - \sin(n-2)x}{\sin x} dx$. Using the trigonometric identity $\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$, we have: $\sin nx - \sin(n-2)x = 2 \cos \left( \frac{nx + nx - 2x}{2} \right) \sin \left( \frac{nx - nx + 2x}{2} \right) = 2 \cos((n-1)x) \sin x$. Substituting this into the integral: $I_n - I_{n-2} = \int \frac{2 \cos((n-1)x) \sin x}{\sin x} dx = 2 \int \cos((n-1)x) dx$. Integrating, we get: $I_n - I_{n-2} = \frac{2 \sin((n-1)x)}{n-1} + C$. To find $I_{n+1} - I_{n-1}$, we replace $n$ with $n+1$: $I_{n+1} - I_{(n+1)-2} = I_{n+1} - I_{n-1} = \frac{2 \sin((n+1-1)x)}{n+1-1} = \frac{2 \sin nx}{n}$. Thus, the correct option is $D$.

If $n$ is a positive integer greater than $1$ and $I_{n}=\int \frac{\sin n x}{\sin x} d x$, then $I_{n+1}-I_{n-1}=$

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