If cos x + cos 2x + ldots + cos nx = frac{A(x | vedclass

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(B) Given, $\cos x + \cos 2x + \ldots + \cos nx = \frac{A(x)}{2 \sin(x/2)}$.Using the sum formula for cosine series, $\sum_{k=1}^n \cos(kx) = \frac{\s

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$\frac{-4n}{2n+1}$

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(B) Given, $\cos x + \cos 2x + \ldots + \cos nx = \frac{A(x)}{2 \sin(x/2)}$.Using the sum formula for cosine series, $\sum_{k=1}^n \cos(kx) = \frac{\sin(nx/2) \cos((n+1)x/2)}{\sin(x/2)}$.Thus, $A(x) = 2 \sin(nx/2) \cos((n+1)x/2)$.Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$A(x) = \sin((n + n/2 + 1/2)x) + \sin((n/2 - (n+1)/2)x) = \sin(\frac{2n+1}{2}x) - \sin(x/2)$.Now, $\int_0^\pi A(x) dx = \int_0^\pi (\sin(\frac{2n+1}{2}x) - \sin(x/2)) dx$.$= [-\frac{2}{2n+1} \cos(\frac{2n+1}{2}x) + 2 \cos(x/2)]_0^\pi$.$= (-\frac{2}{2n+1} \cos(\frac{2n+1}{2}\pi) + 2 \cos(\pi/2)) - (-\frac{2}{2n+1} \cos(0) + 2 \cos(0))$.Since $\cos(\frac{2n+1}{2}\pi) = 0$ and $\cos(\pi/2) = 0$:$= 0 - (-\frac{2}{2n+1} + 2) = \frac{2}{2n+1} - 2 = \frac{2 - 4n - 2}{2n+1} = \frac{-4n}{2n+1}$.

If $\cos x + \cos 2x + \ldots + \cos nx = \frac{A(x)}{2 \sin(x/2)}$, then $\int_0^\pi A(x) dx =$

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