If f(x) = left| begin{array}{ccc} 2 cos^2 x & | vedclass

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(D) First, we evaluate the determinant $f(x)$ by expanding along the third row: $f(x) = \sin x (-\sin x \cos x - 2 \sin^2 x \sin x) - (-\cos x) (-2 \c

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$\pi$

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(D) First, we evaluate the determinant $f(x)$ by expanding along the third row: $f(x) = \sin x (-\sin x \cos x - 2 \sin^2 x \sin x) - (-\cos x) (-2 \cos^3 x - \sin x \sin 2x) + 0$ Simplifying the determinant, we observe that $f(x) = \sin x (- \sin x \cos x - 2 \sin^3 x) + \cos x (-2 \cos^3 x - 2 \sin^2 x \cos x)$ $f(x) = -\sin^2 x \cos x - 2 \sin^4 x - 2 \cos^4 x - 2 \sin^2 x \cos^2 x$ $f(x) = -\sin^2 x \cos x - 2 \sin^2 x (\sin^2 x + \cos^2 x) - 2 \cos^4 x$ $f(x) = -\sin^2 x \cos x - 2 \sin^2 x - 2 \cos^4 x$ Actually, a simpler approach is to note that the determinant evaluates to $f(x) = -1$. Let us re-evaluate: $f(x) = 2 \cos^2 x (0 - \cos^2 x) - \sin 2x (0 + \cos x \sin x) + \sin x (-\sin 2x \cos x - 2 \sin^2 x \sin x)$ $f(x) = -2 \cos^4 x - 2 \sin^2 x \cos^2 x - 2 \sin^2 x \cos^2 x - 2 \sin^4 x$ $f(x) = -2 \cos^2 x (\cos^2 x + \sin^2 x) - 2 \sin^2 x (\cos^2 x + \sin^2 x) = -2(1) - 2(1) = -2$. Since $f(x) = -2$, then $|f(x)| = 2$ and $f'(x) = 0$. Thus, $\int_0^{\frac{\pi}{4}} (2(2) + 5(0)) \, dx = \int_0^{\frac{\pi}{4}} 4 \, dx = 4 	imes \frac{\pi}{4} = \pi$.

If $f(x) = \left| \begin{array}{ccc} 2 \cos^2 x & \sin 2x & \sin x \\ \sin 2x & 2 \sin^2 x & -\cos x \\ \sin x & -\cos x & 0 \end{array} \right|$, then find the value of $\int_0^{\frac{\pi}{4}} (2|f(x)| + 5f'(x)) \, dx$.

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