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If $f (x) = |\begin{matrix} 1 + \sin^{2} x & \cos^{2} x & \sin 2 x \\ \sin^{2} x & 1 + \cos^{2} x & \sin 2 x \\ \sin^{2} x & \cos^{2} x & 1 + \sin 2 x \end{matrix}|$ then $\int_{0}^{\frac{π}{4}} f (x) d x =$

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$\frac{π}{4} - \frac{1}{2}$

$- \frac{π}{2} + \frac{1}{2}$

$\frac{π}{4} - \frac{1}{4}$

$\frac{π}{2} + \frac{1}{2}$

answer is D.

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Detailed Solution

Use Row or columns notations and simplify
we get $f (x) = 2 + \sin 2 x \Rightarrow \int_{0}^{\frac{π}{4}} f (x) d x = 2 \frac{π}{4} - \frac{1}{2} (0 - 1)$

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