If S_n = int_0^{frac{pi}{2}} frac{sin((2n-1)x | vedclass

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(C) Given $S_n = \int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin x} dx$. Then $S_{n+1} = \int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin x} dx$. Now,consider the di

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(C) Given $S_n = \int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin x} dx$. Then $S_{n+1} = \int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin x} dx$. Now,consider the difference: $S_{n+1} - S_n = \int_0^{\pi/2} \frac{\sin((2n+1)x) - \sin((2n-1)x)}{\sin x} dx$. Using the trigonometric identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$,we have: $\sin((2n+1)x) - \sin((2n-1)x) = 2 \cos(2nx) \sin(x)$. Substituting this into the integral: $S_{n+1} - S_n = \int_0^{\pi/2} \frac{2 \cos(2nx) \sin x}{\sin x} dx = \int_0^{\pi/2} 2 \cos(2nx) dx$. Evaluating the integral: $S_{n+1} - S_n = [\frac{2 \sin(2nx)}{2n}]_0^{\pi/2} = [\frac{\sin(2nx)}{n}]_0^{\pi/2}$. Substituting the limits: $S_{n+1} - S_n = \frac{\sin(n\pi) - \sin(0)}{n} = \frac{0 - 0}{n} = 0$,since $n$ is an integer.

If $S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} dx$ and $n$ is an integer,then $S_{n+1} - S_n =$

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