If g(x) = int_{0}^{x} cos 4t , dt,then g(x + | vedclass

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(D) Given $g(x) = \int_{0}^{x} \cos 4t \, dt$. Evaluating the integral,we get $g(x) = \left[ \frac{\sin 4t}{4} \right]_{0}^{x} = \frac{\sin 4x}{4}$. N

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$g(x) + g(\pi)$ and $g(x) - g(\pi)$

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(D) Given $g(x) = \int_{0}^{x} \cos 4t \, dt$. Evaluating the integral,we get $g(x) = \left[ \frac{\sin 4t}{4} \right]_{0}^{x} = \frac{\sin 4x}{4}$. Now,calculate $g(x + \pi) = \frac{\sin 4(x + \pi)}{4} = \frac{\sin(4x + 4\pi)}{4}$. Since $\sin(4x + 4\pi) = \sin 4x$,we have $g(x + \pi) = \frac{\sin 4x}{4} = g(x)$. Also,$g(\pi) = \frac{\sin(4\pi)}{4} = 0$. Since $g(\pi) = 0$,it follows that $g(x) + g(\pi) = g(x) + 0 = g(x)$ and $g(x) - g(\pi) = g(x) - 0 = g(x)$. Thus,both $g(x) + g(\pi)$ and $g(x) - g(\pi)$ are equal to $g(x)$.

If $g(x) = \int_{0}^{x} \cos 4t \, dt$,then $g(x + \pi) = $

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