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If $\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\cos \mathrm{x} & 1 & 0 \\ 1 & 2 \cos \mathrm{x} & 1 \\ 0 & 1 & 2 \cos \mathrm{x}\end{array}\right|$, then $\int_{0}^{\pi / 2} f(x) d x$ is equal to
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Verified Answer
The correct answer is:
$-\frac{1}{3}$
$f(x)=4 \cos ^{3} x-\cos x-2 \cos x=\cos 3 x$
[Expansion of determinant]
$$
\left.\therefore \quad \int_{0}^{\pi / 2} f(x) d x=\frac{\sin 3 x}{3}\right]_{0}^{\pi / 2}=-\frac{1}{3}
$$
[Expansion of determinant]
$$
\left.\therefore \quad \int_{0}^{\pi / 2} f(x) d x=\frac{\sin 3 x}{3}\right]_{0}^{\pi / 2}=-\frac{1}{3}
$$
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