Given,f(x)=0x2−sin ⁡xcos ⁡x−2sin ⁡x−x201−2x2−cos⁡ x2x−10' then ∫f(x)dx is equal to

A
$\frac{x^{3}}{3} - x^{2} \sin x + \sin 2 x + C$
B
$\frac{x^{3}}{3} - x^{2} \sin x - \cos 2 x + C$
C
$\frac{x^{3}}{3} - x^{2} \cos x - \cos 2 x + C$
D
None of the above

Solution:

We have,

$\Rightarrow$ $\begin{aligned} f (x) = |\begin{array}{ccl} 0 & x^{2} - \sin x & \cos x - 2 \\ \sin x - x^{2} & 0 & 1 - 2 x \\ 2 - \cos x & 2 x - 1 & 0 \end{array}| \\ f (x) = |\begin{array}{ccc} 0 & \sin x - x^{2} & 2 - \cos x \\ x^{2} - \sin x & 0 & 2 x - 1 \\ \cos x - 2 & 1 - 2 x & 0 \end{array}| \end{aligned}$

[interchanging rows and columns]

$\Rightarrow$ $f (x) = (- 1)^{3} |\begin{array}{ccc} 0 & x^{2} - \sin x & \cos x - 2 \\ \sin x - x^{2} & 0 & 1 - 2 x \\ 2 - \cos x & 2 x - 1 & 0 \end{array}|$

[taking (-1) common from each column]

$\begin{array}{l} \Rightarrow f (x) = - f (x) \Rightarrow f (x) = 0 \\ \Rightarrow \int f (x) dx = c \end{array}$

Given, $f (x) = |\begin{array}{ccc} 0 & x^{2} - \sin x & \cos x - 2 \\ \sin x - x^{2} & 0 & 1 - 2 x \\ 2 - \cos x & 2 x - 1 & 0 \end{array}|$ ' then $\int f (x) dx$ is equal to

Solution:

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