Question

If $f\left(x\right)=\left|\begin{array}{ccc}{\sec }^{2}x& 1& 1\\ {\cos }^{2}x& {\cos }^{2}x& {\text{cosec}}^{2}x\\ 1& {\cos }^{2}x& {\cot }^{2}x\end{array}\right|$, then ${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx=$

• A. $\frac{1}{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{8}$
• D. $\frac{3\pi }{16}$

Answer 1

The correct option is D $\frac{3\pi }{16}$

$f\left(x\right)=\left|\begin{array}{ccc}{\sec }^{2}x& 1& 1\\ {\cos }^{2}x& {\cos }^{2}x& {\text{cosec}}^{2}x\\ 1& {\cos }^{2}x& {\cot }^{2}x\end{array}\right|$
$={\sec }^{2}x({\cos }^{2}x.{\cot }^{2}x-{\cos }^{2}x.{\text{cosec}}^{2}x)-1({\cos }^{2}x.{\cot }^{2}x-{\text{cosec}}^{2}x)+1({\cos }^{2}x.{\cos }^{2}x-{\cos }^{2}x)$
$={\cos }^{4}x$
Therefore,
${\int }_0^{\frac{\pi }{2}}f\left(x\right)dx={\int }_0^{\frac{\pi }{2}}{\cos }^{4}xdx={\left[\frac{1}{32}(12x+8\sin 2x+\sin 4x)\right]}_0^{\frac{\pi }{2}}=\frac{1}{32}\left(6\pi \right)=\frac{3\pi }{16}$