, then equals

Engineering

Mathematics

Introduction to Determinants

Question

$∣ f (x) ∣ = [\begin{array}{ccc} sinx & cosecx & tanx \\ secx & xsinx & xtanx \\ x^{2} - 1 & cosx & x^{2} + 1 \end{array}]$ , then $\int_{- a}^{a} ∣ f (x) ∣ d$ equals

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$∣ f (x) ∣ = [\begin{array}{ccc} \sin x & \csc x & \tan x \\ \sec x & xsin x & xtan x \\ x^{2} - 1 & - \sin x & x^{2} + 1 \end{array}]$
$\int_{- a}^{a} ∣ f (x) ∣ dx = [\begin{array}{ccc} \int_{- a}^{a} \sin xdx & \int_{- a}^{a} \csc xdx & \int_{- a}^{a} \tan xdx \\ \int_{- a}^{a} \sec xdx & \int_{- a}^{a} xsin xdx & \int_{- a}^{a} xtan xdx \\ \int_{- a}^{a} (x^{2} - 1) dx & - \int_{- a}^{a} \sin xdx & \int_{- a}^{a} (x^{2} + 1) dx \end{array}]$
$\int_{- a}^{a} \sin xdx = 0$ , since sinx is a odd function

similarly

\int_{- a}^{a} \csc xdx = 0

and

\int_{- a}^{a} \tan xdx = 0

, since cscx and tanx are odd functions

= [\begin{array}{ccc} 0 & 0 & 0 \\ \int_{- a}^{a} \sec xdx & \int_{- a}^{a} xsin xdx & \int_{- a}^{a} xtan xdx \\ \int_{- a}^{a} (x^{2} - 1) dx & - \int_{- a}^{a} \sin xdx & \int_{- a}^{a} (x^{2} + 1) dx \end{array}]

= 0

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