Question

$f(x)=\left|\begin{array}{ccc}secx& cosx& cose{c}^{2}x\\ co{s}^{2}x& co{s}^{2}x& cose{c}^{2}x\\ 1& co{s}^{2}x& co{s}^{2}x\end{array}\right|$
Then, ${\int }_0^{\pi /2}f(x)dx=....$

Answer 1

Given, $f(x)=\left|\begin{array}{ccc}secx& cosx& cosecx.cotx+se{c}^{2}x\\ co{s}^{2}x& co{s}^{2}x& cose{c}^{2}x\\ 1& co{s}^{2}x& co{s}^{2}x\end{array}\right|$
Applying $R_3\to \frac{1}{cosx}{R}_3,$
$f(x)=cosx\left|\begin{array}{ccc}secx& cosx& cosecx.cotx+se{c}^{2}x\\ co{s}^{2}x& co{s}^{2}x& cose{c}^{2}x\\ secx& co{s}^{2}x& cosx\end{array}\right|$
Applying $R_1\to R_1-R_3\Rightarrow f(x)$
$=cosx\left|\begin{array}{ccc}0& 0& cosecx.cotx+se{c}^{2}x-cosx\\ co{s}^{2}x& co{s}^{2}x& cose{c}^{2}x\\ secx& co{s}^{2}x& cosx\end{array}\right|$
$=(cosecx.cotx+se{c}^{2}x-cosx).(co{s}^{3}x-cosx).cosx$
$=-\left[\frac{si{n}^{2}x+co{s}^{2}x-co{s}^{3}x.si{n}^{2}x}{si{n}^{2}x.co{s}^{2}x}\right].co{s}^{2}x.si{n}^{2}x$
$=-si{n}^{2}x-co{s}^{3}x(1-si{n}^{2}x)=-si{n}^{2}x-co{s}^{5}x$
$\therefore {\int }_0^{\pi /2}f(x)dx=-{\int }_0^{\pi /2}(si{n}^2+co{s}^{5}x)dx$
$\therefore {\int }_0^{\pi /2}f(x)dx=-{\int }_0^{\pi /2}(si{n}^{2}x+co{s}^{5}x)dx$
$[\because {\int }_0^{\pi /2}si{n}^{m}x.co{s}^{n}xdx=\frac{\sqrt{\frac{m+1}{2}}\sqrt{\frac{n+2}{2}}}{2\sqrt{\frac{m+n+2}{2}}}]$ ${\int }_0^{\pi /2}f(x)dx=-\{\frac{\sqrt{\frac{5}{2}},\sqrt{\frac{1}{2}}}{2\sqrt{2}}+\frac{\sqrt{\frac{2}{2}}\sqrt{\frac{1}{3}}}{2\sqrt{\frac{7}{2}}}\}$
$=-\{\frac{1/2.\pi }{2}+\frac{2\sqrt{\pi }}{2.\frac{5}{2}.\frac{3}{2}.\frac{1}{2}.\sqrt{\pi }}\}=-\{\frac{\pi }{4}+\frac{8}{15}\}=-\left(\frac{15\pi +32}{60}\right)$