Question

If $f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot xcosecx\\ {\cos }^{2}x& {\cos }^{2}x& cose{c}^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$ then ${\int }_{-\pi }^{\pi }f\left(x\right)dx$ equals

• A. $0$
• B. $-\pi$
• C. $\pi$
• D. None of these

Answer 1

The correct option is B $-\pi$

$f\left(x\right)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot xco\sec x\\ {\cos }^{2}x& {\cos }^{2}x& co{\sec }^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$
Taking out common $\cos x$ from $C_2$
$=\cos x\left|\begin{array}{ccc}\sec x& 1& {\sec }^{2}x+\cot xco\sec x\\ {\cos }^{2}x& \cos x& co{\sec }^{2}x\\ 1& \cos x& {\cos }^{2}x\end{array}\right|$
Applying $C_2\to C_2-\cos x{C}_1$
$=\cos x\left|\begin{array}{ccc}\sec x& 0& {\sec }^{2}x+\cot xco\sec x\\ {\cos }^{2}x& \cos x-{\cos }^{3}x& co{\sec }^{2}x\\ 1& 0& {\cos }^{2}x\end{array}\right|$

$f\left(x\right)=\cos x(\cos x-{\cos }^{3}x)(\cos x-\frac{1}{{\cos }^{2}x}-\frac{\cos x}{{\sin }^{2}x})$
$f\left(x\right)=-{\cos }^{5}x-{\sin }^{2}x$
$\therefore {\int }_{-\pi }^{\pi }f\left(x\right)dx={\int }_{-\pi }^{\pi }-{\cos }^{5}xdx-{\int }_{-\pi }^{\pi }{\sin }^{2}xdx$ (because $f\left(x\right)$ is even in both integrals and

${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx)$
$=-2{\int }_0^{\pi }{\cos }^{5}xdx-2{\int }_0^{\pi }{\sin }^{2}xdx$
$=0-2\times 2{\int }_0^{\pi /2}{\sin }^{2}xdx$
Using ${\int }_0^{2a}f\left(x\right)dx=\left\{\begin{array}{cc}2{\int }_0^{a}f\left(x\right)dx& iff(2a-x)=f\left(x\right)\\ 0& iff(2a-x)=-f\left(x\right)\end{array}\right\}$
$=-4\times \frac{1}{2}\times \frac{\pi }{2}=-\pi$