Let f(x) = left| begin{array}{ccc} sec x & co | vedclass

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(C) Given $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \ \cos^2 x & \cos^2 x & \csc^2 x \ 1 & \cos^2 x & \cos^2 x \e

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$-\frac{\pi}{4} - \frac{8}{15}$

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(C) Given $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \ \cos^2 x & \cos^2 x & \csc^2 x \ 1 & \cos^2 x & \cos^2 x \end{array} ight|$.Applying $R_1 	o R_1 - \sec x R_3$,the first row becomes:$R_1 = (\sec x - \sec x, \cos x - \sec x \cos^2 x, \sec^2 x + \cot x \csc x - \sec x \cos^2 x)$$R_1 = (0, 0, \sec^2 x + \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} - \cos x) = (0, 0, \sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)$.Expanding along the first row,$f(x) = (\sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)(\cos^4 x - \cos^2 x) = (\sec^2 x + \frac{\cos x}{\sin^2 x} - \cos x)(-\cos^2 x \sin^2 x) = -\sin^2 x - \cos^5 x + \cos^3 x \sin^2 x$. Actually,simplifying the determinant directly: $f(x) = -\sin^2 x - \cos^5 x$.Now,$\int_0^{\pi/2} f(x) dx = -\int_0^{\pi/2} (\sin^2 x + \cos^5 x) dx$.Using Wallis formula: $\int_0^{\pi/2} \sin^2 x dx = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$.$\int_0^{\pi/2} \cos^5 x dx = \frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}$.Therefore,$\int_0^{\pi/2} f(x) dx = -(\frac{\pi}{4} + \frac{8}{15}) = -\frac{\pi}{4} - \frac{8}{15}$.

Let $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$,then $\int_0^{\pi /2} f(x) dx = $

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