Question

Let $f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\csc x\\ {\cos }^{2}x& {\cos }^{2}x& {\csc }^{2}x\\ 1& {\cos }^{2}x& {\csc }^{2}x\end{array}\right|$ then the value of ${\int }_{\pi /4}^{\pi /2}f\left(x\right)dx$ is?

• A. $0$
• B. $\pi /48$
• C. $-\frac{\pi }{2}-\frac{\pi }{15\sqrt{2}}$
• D. None of the above

Answer 1

The correct option is D None of the above

Applying $R_2\to R_2-R_3$
$f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\csc x\\ -{\sin }^{2}x& 0& 0\\ 1& {\cos }^{2}x& {\csc }^{2}x\end{array}\right|$
$=-(-{\sin }^{2}x)=\left|\begin{array}{cc}\cos x& {\sec }^{2}x+\cot x\csc x\\ {\cos }^{2}x& {\csc }^{2}x\end{array}\right|$
$={\sin }^{2}x[\cos x{\csc }^{2}x-{\cos }^{2}x({\sec }^{2}x+\cot x\csc x)]$
$\cos x-{\sin }^{2}x-{\cos }^{2}x=\cos x{\sin }^{2}x-\frac{1}{2}(1-{\cos }^{2}x)$
Thus ,
${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}f\left(x\right)dx={[\frac{1}{3}{\sin }^{3}x-\frac{1}{2}x+\frac{1}{4}{\sin }^{2}x]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$
$=\frac{1}{3}-\frac{\pi }{4}-\frac{1}{3}\cdot \frac{1}{2\sqrt{2}}+\frac{\pi }{8}-\frac{1}{4}$
$=\frac{1}{12}-\frac{\pi }{8}-\frac{1}{6\sqrt{2}}$