Mathematics

If $$f(x) \begin{vmatrix} x & cosx & e^{ { x^{ 2 } } } \\ sin\quad x & x^{ 2 } & sec\quad x \\ tan\quad x & x^ 4 & 2x^ 2 \end{vmatrix}$$ then $$\displaystyle \int_{-\pi/2}^{\pi/2}f(x)dx=$$


ANSWER

$$0$$


SOLUTION
$$f(x) = \begin{vmatrix}  x& \cos x & e^{x^2}\\ \sin x & x^2 &\sec x\\\tan x& x^4 & 2x^2  \end{vmatrix}$$

$$=x[2x^4-x^4\sec x]-\cos x[2x^2\sin x +\sec x \tan x]+e^{x^2}[x^4\sin x - x^2\tan x]$$

$$=x^5[2-\sec x]-2x^2\sin x \cos x + \tan x+e^{x^2}x^4\sin x -e^{x^2}\tan x$$

$$\displaystyle \int_{-\pi/2}^{\pi/2} [x^5[2-\sec x]-2x^2 \sin x \cos x + \tan x + e^{x^2}x^4\sin x - e^{x^2}x^2\tan x]$$

$$\left[ \dfrac{2x^6}{6}= x^5 \sec x - 2x^2\sin x \cos x - \tan x + e^{x^2}x^4\sin x - e^{x^2}x^2\tan x\right]$$

$$=0$$
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Single Correct Medium Published on 17th 09, 2020
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