intlimits_0^pi {frac{{sin left( {n + frac{1}{ | vedclass

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(C) We know that $2\sin \frac{x}{2} \left( {\frac{1}{2} + \cos x + \cos 2x + \dots + \cos nx} \right) = \sin \frac{x}{2} + \sum_{k=1}^n 2\sin \frac{x}

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$\pi $

Vedclass · Answer

(C) We know that $2\sin \frac{x}{2} \left( {\frac{1}{2} + \cos x + \cos 2x + \dots + \cos nx} \right) = \sin \frac{x}{2} + \sum_{k=1}^n 2\sin \frac{x}{2} \cos kx$. Using the identity $2\sin A \cos B = \sin(A+B) + \sin(A-B)$,we get: $2\sin \frac{x}{2} \cos kx = \sin \left( k + \frac{1}{2} \right)x - \sin \left( k - \frac{1}{2} \right)x$. Summing this from $k=1$ to $n$,we get a telescoping series: $\sin \frac{x}{2} + \left( \sin \frac{3x}{2} - \sin \frac{x}{2} \right) + \left( \sin \frac{5x}{2} - \sin \frac{3x}{2} \right) + \dots + \left( \sin \left( n + \frac{1}{2} \right)x - \sin \left( n - \frac{1}{2} \right)x \right) = \sin \left( n + \frac{1}{2} \right)x$. Thus,$\frac{\sin \left( n + \frac{1}{2} \right)x}{\sin \frac{x}{2}} = 1 + 2\cos x + 2\cos 2x + \dots + 2\cos nx$. Now,integrate from $0$ to $\pi$: $\int_0^\pi \left( 1 + 2\sum_{k=1}^n \cos kx \right) dx = \left[ x + 2\sum_{k=1}^n \frac{\sin kx}{k} \right]_0^\pi$. Since $\sin(k\pi) = 0$ for all integers $k$,the integral evaluates to $\pi - 0 = \pi$.

$\int\limits_0^\pi {\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{{\sin \frac{x}{2}}}} \,dx$,$(n \in N)$ equals

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