The value of the definite integral,intlimits_ | vedclass

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(B) Let $I_n = \int\limits_0^{\frac{\pi }{2}} \frac{\sin(nx)}{\sin x} dx$. We know that $\sin(nx) - \sin((n-2)x) = 2 \cos((n-1)x) \sin x$. Therefore,$

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$\frac{\pi }{2}$

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(B) Let $I_n = \int\limits_0^{\frac{\pi }{2}} \frac{\sin(nx)}{\sin x} dx$. We know that $\sin(nx) - \sin((n-2)x) = 2 \cos((n-1)x) \sin x$. Therefore,$\frac{\sin(nx)}{\sin x} - \frac{\sin((n-2)x)}{\sin x} = 2 \cos((n-1)x)$. Integrating both sides from $0$ to $\frac{\pi}{2}$: $I_n - I_{n-2} = \int\limits_0^{\frac{\pi}{2}} 2 \cos((n-1)x) dx = \left[ \frac{2 \sin((n-1)x)}{n-1} \right]_0^{\frac{\pi}{2}}$. For $n=5$: $I_5 - I_3 = \left[ \frac{2 \sin(4x)}{4} \right]_0^{\frac{\pi}{2}} = \frac{1}{2} (\sin(2\pi) - \sin(0)) = 0$. Thus,$I_5 = I_3$. Similarly,$I_3 - I_1 = \left[ \frac{2 \sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = [\sin(2x)]_0^{\frac{\pi}{2}} = \sin(\pi) - \sin(0) = 0$. Thus,$I_3 = I_1$. Finally,$I_1 = \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x} dx = \int\limits_0^{\frac{\pi}{2}} 1 dx = \frac{\pi}{2}$. Hence,$I_5 = I_3 = I_1 = \frac{\pi}{2}$.

The value of the definite integral,$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin 5x}}{{\sin x}}\,dx} $ is

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