ધારો કે f(x) = left| begin{array}{ccc} sec x | vedclass

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(C) આપેલ છે કે $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \ \cos^2 x & \cos^2 x & \csc^2 x \ 1 & \cos^2 x & \cos^2

Vedclass · Accepted Answer

$-\frac{\pi}{4} - \frac{8}{15}$

Vedclass · Answer

(C) આપેલ છે કે $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \ \cos^2 x & \cos^2 x & \csc^2 x \ 1 & \cos^2 x & \cos^2 x \end{array} ight|$.$R_1 	o R_1 - \sec x R_3$ પ્રક્રિયા લાગુ પાડતા,નિશ્ચાયકનું સાદું રૂપ $f(x) = -\sin^2 x - \cos^5 x$ મળે છે.હવે,$\int_0^{\pi/2} f(x) dx = -\int_0^{\pi/2} (\sin^2 x + \cos^5 x) dx$.વોલિસના સૂત્રનો ઉપયોગ કરતા: $\int_0^{\pi/2} \sin^2 x dx = \frac{\pi}{4}$ અને $\int_0^{\pi/2} \cos^5 x dx = \frac{8}{15}$.તેથી,$\int_0^{\pi/2} f(x) dx = -(\frac{\pi}{4} + \frac{8}{15}) = -\frac{\pi}{4} - \frac{8}{15}$.

ધારો કે $f(x) = \left| \begin{array}{ccc} \sec x & \cos x & \sec^2 x + \cot x \csc x \\ \cos^2 x & \cos^2 x & \csc^2 x \\ 1 & \cos^2 x & \cos^2 x \end{array} \right|$,તો $\int_0^{\pi /2} f(x) dx = $

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