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Question Number 59170 by 2772639291927 last updated on 05/May/19

Iff(x)= determinant (((sec x),(cos x),(sec^2 x+cosec x cot x)),((cos^2 x),(cos^2 x),(          cosec^2 x)),((   1),(cos^2 x),(          cos^2 x)))  then ∫_( 0) ^(π/2)  f(x) dx =

$$\mathrm{If}{f}\left({x}\right)=\begin{vmatrix}{\mathrm{sec}\:{x}}&{\mathrm{cos}\:{x}}&{\mathrm{sec}^{\mathrm{2}} {x}+\mathrm{cosec}\:{x}\:\mathrm{cot}\:{x}}\\{\mathrm{cos}^{\mathrm{2}} {x}}&{\mathrm{cos}^{\mathrm{2}} {x}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{cosec}^{\mathrm{2}} {x}}\\{\:\:\:\mathrm{1}}&{\mathrm{cos}^{\mathrm{2}} {x}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{cos}^{\mathrm{2}} {x}}\end{vmatrix} \\ $$$$\mathrm{then}\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:{f}\left({x}\right)\:{dx}\:= \\ $$

Answered by MJS last updated on 05/May/19

if f(x)=det (above matrix):  f(x)=−((1/(16))cos 5x +(5/(16))cos 3x −(1/2)cos 2x +(5/8)cos x +(1/2))  F(x)=∫f(x)dx=−((1/(80))sin 5x +(5/(48))sin 3x −(1/4)sin 2x +(5/8)sin x +(1/2)x)w  F((π/2))−F(0)=−(π/2)−(8/(15))

$$\mathrm{if}\:{f}\left({x}\right)=\mathrm{det}\:\left(\mathrm{above}\:\mathrm{matrix}\right): \\ $$$${f}\left({x}\right)=−\left(\frac{\mathrm{1}}{\mathrm{16}}\mathrm{cos}\:\mathrm{5}{x}\:+\frac{\mathrm{5}}{\mathrm{16}}\mathrm{cos}\:\mathrm{3}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\:+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{cos}\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${F}\left({x}\right)=\int{f}\left({x}\right){dx}=−\left(\frac{\mathrm{1}}{\mathrm{80}}\mathrm{sin}\:\mathrm{5}{x}\:+\frac{\mathrm{5}}{\mathrm{48}}\mathrm{sin}\:\mathrm{3}{x}\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{x}\:+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{sin}\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}{x}\right){w} \\ $$$${F}\left(\frac{\pi}{\mathrm{2}}\right)−{F}\left(\mathrm{0}\right)=−\frac{\pi}{\mathrm{2}}−\frac{\mathrm{8}}{\mathrm{15}} \\ $$

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