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Question

If $$f(x)=\begin{vmatrix} x & cosx & { e }^{ \left| x \right|  } \\ sinx & x^{ 2 } & secx \\ tanx & 1 & 2 \end{vmatrix}$$ then the value of $$\displaystyle \int_{- \pi / 2}^{\pi/2}f(x) dx$$ is equal to .................


Solution

$$f(x)=\begin{vmatrix} x & cosx & { e }^{ \left| x \right|  } \\ sinx & x^{ 2 } & secx \\ tanx & 1 & 2 \end{vmatrix}\\ f(-x)=\begin{vmatrix} -x & cos-x & { e }^{ \left| -x \right|  } \\ sin-x & \left( -x \right) ^{ 2 } & sec-x \\ tan-x & 1 & 2 \end{vmatrix}=\begin{vmatrix} -x & cosx & { e }^{ \left| x \right|  } \\ -sinx & \left( x \right) ^{ 2 } & secx \\ -tanx & 1 & 2 \end{vmatrix}=-\begin{vmatrix} x & cosx & { e }^{ \left| x \right|  } \\ sinx & x^{ 2 } & secx \\ tanx & 1 & 2 \end{vmatrix}$$
As $$f(x)+f(-x)=0$$
Then $$\int _{ -\pi /2 }^{ \pi /2 } f(x)=0$$

Mathematics

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