Question

State true or false:

If $f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\cos ecx\\ {\cos }^{2}x& {\cos }^{2}x& \cos e{c}^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$, then ${\int }_0^{\pi /2}f\left(x\right)dx=-\left(\frac{15\pi +32}{60}\right)$.

• A. True
• B. False

Answer 1

The correct option is A True

$f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\csc x\\ {\cos }^{2}x& {\cos }^{2}x& {\csc }^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$

Applying $R_3\to \frac{1}{\cos x}{R}_3$

$f\left(x\right)=\cos x\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\csc x\\ {\cos }^{2}x& {\cos }^{2}x& {\csc }^{2}x\\ \sec x& \cos x& \cos x\end{array}\right|$

Again applying $R_1\to R_1-R_3$

$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}0& 0& {\sec }^{2}x+\cot x\csc x-\cos x\\ {\cos }^{2}x& {\cos }^{2}x& {\csc }^{2}x\\ \sec x& \cos x& \cos x\end{array}\right|$

$=({\sec }^{2}x+\cot x\csc x-\cos x)({\cos }^{3}x-\cos x)\cos x=-\left[\frac{{\sin }^{2}x+{\cos }^{3}x-{\cos }^{3}x{\sin }^{2}x}{{\sin }^{2}x{\cos }^{2}x}\right]{\cos }^{2}x{\sin }^{2}x=-{\sin }^{2}x-{\cos }^{3}x(1-{\sin }^{2}x)=-{\sin }^{2}x-{\cos }^{5}x{\int }_0^{\pi /2}f\left(x\right)dx=-{\int }_0^{\pi /2}({\sin }^{2}x+{\cos }^{5}x)dx=-(\frac{\pi }{4}+\frac{8}{15})=-\left(\frac{15\pi +32}{60}\right)$