# Integration of Determinant

Let $\large \Delta (x) = \left| \begin{array}{ccc} f(x) & g(x) & h(x) \\ a & b & c \\ l & m & n \end{array} \right|$

where a, b, c, l, m and n are constants.

$\large \int_{a}^{b}\Delta (x) = \left| \begin{array}{ccc} \int_{a}^{b} f(x) & \int_{a}^{b} g(x) & \int_{a}^{b} h(x) \\ a & b & c \\ l & m & n \end{array} \right|$

Note:
∎ If the elements of more than one column or rows are functions of x then the integration can be done only after evaluation / expansion of the determinant.

Illustration :

Let $\large f(x) = \left| \begin{array}{ccc} secx & cosx & sec^2 x + cotx cosecx \\ cos^2 x & cos^2 x & cosec^2 x \\ 1 & cos^2 x & cos^2 x \end{array} \right|$

Prove that : $\large \int_{0}^{\pi/2} f(x) dx = – (\frac{\pi}{4} + \frac{8}{15})$

Solution: Operate R1 → R1 – secx R3

$\large f(x) = \left| \begin{array}{ccc} 0 & 0 & sec^2 x + cotx cosecx – cosx\\ cos^2 x & cos^2 x & cosec^2 x \\ 1 & cos^2 x & cos^2 x \end{array} \right|$

$\large = ( sec^2 x + cotx cosecx – cosx ) (cos^4 x – cos^2 x)$

$\large f(x) = (1+ \frac{cos^3 x}{sin^2 x} – cos^3 x) (cos^2 x – 1)$

$\large = – sin^2 x \frac{sin^2 x + cos^3 x – cos^3 x sin^2 x}{sin^2 x}$

$\large f(x) = -(sin^2 x + cos^5 x)$

$\large \int_{0}^{\pi/2} f(x) dx = – \int_{0}^{\pi/2} (sin^2 x + cos^5 x) dx$

$\large = -(\frac{1}{2}.\frac{\pi}{2} + \frac{4 . 2}{5.3})$

$\large = -(\frac{\pi}{4} + \frac{8}{15})$

Exercise :( i) $\large U_n = \int_{0}^{\pi/2} \frac{1-cos2nx}{1-cos2x} dx$

Then show that: $\large \Delta = \left| \begin{array} {ccc} \pi/2 & U_2 & U_3 \\ U_4 & U_5 & U_6 \\ U_7 & U_8 & U_9 \end{array}\right| = 0$

(ii) If $\large f(x) = \left| \begin{array} {ccc} 2cos^2 x & sin2x & -sinx \\ sin2x & 2sin^2 x & cosx \\ sinx & -cosx & 0 \end{array}\right|$

Then prove that: $\large \int_{0}^{\pi/2} (f(x) + f'(x)) dx = \pi$

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