Question

$f(x)=\begin {vmatrix} sec x & cos x & cosec^2x \\ cos^2x & cos^2x & cosec^2x \\ 1 & cos^2x & cos^2x \end {vmatrix}$
Then, $\int_0^{\pi/2} \ f(x) \ dx=....$

Answer 1

Given, $f(x)=\begin {vmatrix} sec x & cosx & cosecx.cotx+sec^2x \\ cos^2x & cos^2x & cosec^2x \\ 1 & cos^2x & cos^2x \end {vmatrix}$
Applying $R_3 \rightarrow \ \frac{1}{cos x}R_3,$
$f(x)=cos \ x\begin {vmatrix} sec x & cosx & cosecx.cotx+sec^2x \\ cos^2x & cos^2x & cosec^2x \\ secx & cos^2x & cos x \end {vmatrix}$
Applying $R_1 \rightarrow \ R_1 -R_3 \ \ \ \Rightarrow \ \ f(x)$
$=cos \ x\begin {vmatrix} 0 & 0 & cosecx.cotx+sec^2x-cos x \\ cos^2x & cos^2x & cosec^2x \\ secx & cos^2x & cos x \end {vmatrix}$
$=( cosecx.cotx+sec^2x-cos x).(cos^3x-cosx).cos x$
$=-\bigg[\frac{sin^2x+cos^2x-cos^3x.sin^2x}{sin^2x.cos^2x}\bigg] .cos^2x.sin^2x$
$=-sin^2x-cos^3x(1-sin^2x)=-sin^2x-cos^5x$
$\therefore \ \ \int_0^{\pi/2} \ f(x) dx \ =-\int_0^{\pi/2}(sin^2+cos^5x) dx$
$\therefore \ \ \ \int_0^{\pi/2} f(x) dx =-\int_0^{\pi/2} (sin^2 x+cos^5 x) dx$
$\Bigg[\because \ \int_0^{\pi/2} \ sin^m x.cos^n x dx=\frac{\sqrt{\frac{m+1}{2}}\sqrt{\frac{n+2}{2}}}{2\sqrt{\frac{m+n+2}{2}}}\Bigg]$ $\int_0^{\pi/2}f(x)dx=-\Bigg \{\frac{\sqrt{\frac{5}{2}},\sqrt{\frac{1}{2}}}{2\sqrt 2}+\frac{\sqrt{\frac{2}{2}}\sqrt{\frac{1}{3}}}{2\sqrt{\frac{7}{2}}}\Bigg\}$
$=-\Bigg \{ \frac{1/2.\pi}{2}+\frac{2\sqrt \pi}{2.\frac{5}{2}.\frac{3}{2}.\frac{1}{2}.\sqrt \pi}\Bigg \}= -\bigg\{\frac{\pi}{4}+\frac{8}{15}\bigg\} =-\bigg(\frac{15\pi+32}{60}\bigg)$