If int e^x(sin^2 2x - 8 cos 4x) dx = e^x f(x) | vedclass

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(B) We are given the integral $I = \int e^x(\sin^2 2x - 8 \cos 4x) dx$. Using the identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$,we substitute this into

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(B) We are given the integral $I = \int e^x(\sin^2 2x - 8 \cos 4x) dx$. Using the identity $\sin^2 2x = \frac{1 - \cos 4x}{2}$,we substitute this into the integral: $I = \int e^x(\frac{1 - \cos 4x}{2} - 8 \cos 4x) dx = \int e^x(\frac{1}{2} - \frac{1}{2} \cos 4x - 8 \cos 4x) dx = \int e^x(\frac{1}{2} - \frac{17}{2} \cos 4x) dx$. This simplifies to $I = \frac{1}{2} \int e^x dx - \frac{17}{2} \int e^x \cos 4x dx$. Using the standard formula $\int e^{ax} \cos(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$,we have $\int e^x \cos 4x dx = \frac{e^x}{1^2 + 4^2} (\cos 4x + 4 \sin 4x) = \frac{e^x}{17} (\cos 4x + 4 \sin 4x)$. Substituting this back: $I = \frac{1}{2} e^x - \frac{17}{2} \cdot \frac{e^x}{17} (\cos 4x + 4 \sin 4x) + c = e^x (\frac{1}{2} - \frac{1}{2} \cos 4x - 2 \sin 4x) + c$. Thus,$f(x) = \frac{1}{2} - \frac{1}{2} \cos 4x - 2 \sin 4x$. Evaluating at $x = \frac{\pi}{4}$: $f(\frac{\pi}{4}) = \frac{1}{2} - \frac{1}{2} \cos(\pi) - 2 \sin(\pi) = \frac{1}{2} - \frac{1}{2}(-1) - 2(0) = \frac{1}{2} + \frac{1}{2} = 1$.

If $\int e^x(\sin^2 2x - 8 \cos 4x) dx = e^x f(x) + c$,then $f(\frac{\pi}{4}) = $

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