If I_n = int_{0}^{pi} frac{sin(nx)}{sin(x)} d | vedclass

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(B) Consider the difference $I_{n+2} - I_n = \int_{0}^{\pi} \frac{\sin((n+2)x) - \sin(nx)}{\sin(x)} dx$. Using the formula $\sin(A) - \sin(B) = 2 \cos

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$5\pi$

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(B) Consider the difference $I_{n+2} - I_n = \int_{0}^{\pi} \frac{\sin((n+2)x) - \sin(nx)}{\sin(x)} dx$. Using the formula $\sin(A) - \sin(B) = 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$,we get $\sin((n+2)x) - \sin(nx) = 2 \cos((n+1)x) \sin(x)$. Thus,$I_{n+2} - I_n = \int_{0}^{\pi} 2 \cos((n+1)x) dx = [\frac{2 \sin((n+1)x)}{n+1}]_{0}^{\pi} = 0$. This implies $I_n = I_{n+2}$. For $n=1$,$I_1 = \int_{0}^{\pi} 1 dx = \pi$. For $n=2$,$I_2 = \int_{0}^{\pi} \frac{\sin(2x)}{\sin(x)} dx = \int_{0}^{\pi} 2 \cos(x) dx = [2 \sin(x)]_{0}^{\pi} = 0$. Since $I_n = I_{n+2}$,all odd terms $I_1, I_3, \dots, I_9$ are equal to $\pi$,and all even terms $I_2, I_4, \dots, I_{10}$ are equal to $0$. Therefore,$\sum_{n=1}^{10} I_n = (I_1 + I_3 + I_5 + I_7 + I_9) + (I_2 + I_4 + I_6 + I_8 + I_{10}) = 5(\pi) + 5(0) = 5\pi$.

If $I_n = \int_{0}^{\pi} \frac{\sin(nx)}{\sin(x)} dx$,then the value of $\sum_{n=1}^{10} I_n$ is-

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