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

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(A) First,we simplify the determinant $f(x)$.Expanding along the third row: $f(x) = \sin x [\sin(2x) \cos x - 2\sin^2 x(-\sin x)] - (-\cos x) [2\cos^2

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$\pi$

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(A) First,we simplify the determinant $f(x)$.Expanding along the third row: $f(x) = \sin x [\sin(2x) \cos x - 2\sin^2 x(-\sin x)] - (-\cos x) [2\cos^2 x \cos x - (-\sin x)(\sin 2x)] + 0$.$f(x) = \sin x [2\sin x \cos^2 x + 2\sin^3 x] + \cos x [2\cos^3 x + 2\sin^2 x \cos x]$.$f(x) = 2\sin^2 x \cos^2 x + 2\sin^4 x + 2\cos^4 x + 2\sin^2 x \cos^2 x$.$f(x) = 2(\sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x) = 2(\sin^2 x + \cos^2 x)^2 = 2(1)^2 = 2$.Since $f(x) = 2$,then $f'(x) = 0$.Therefore,$\int_{0}^{\frac{\pi}{2}} [f(x) + f'(x)] dx = \int_{0}^{\frac{\pi}{2}} [2 + 0] dx = \int_{0}^{\frac{\pi}{2}} 2 dx = [2x]_{0}^{\frac{\pi}{2}} = 2(\frac{\pi}{2} - 0) = \pi$.

Let $f(x) = \left| \begin{array}{ccc} 2\cos^2 x & \sin(2x) & -\sin x \\ \sin(2x) & 2\sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array} \right|$. Then,evaluate $\int_{0}^{\frac{\pi}{2}} [f(x) + f'(x)] dx$.

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