Question

Let $f(x)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\cdot \csc x\\ {\cos }^{2}x& {\cos }^{2}x& {\csc }^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$ then ${\int }_{-\pi }^{\pi }f(x)dx$ has the value equal to

• A. $\frac{-\pi }{4}$
• B. $\frac{-\pi }{2}$
• C. $-\pi$
• D. $-2\pi$

Answer 1

Answer: (C)

$R_3\to R_3-(\cos x)R_1$
$f(x)=\left|\begin{array}{ccc}\sec x& \cos x& \left({\sec }^{2}x+\cot x\cdot \mathrm{cosec}x\right)\\ {\cos }^{2}x& {\cos }^{2}x& \mathrm{cosec}2\\ 0& 0& {\cos }^{2}x-\sec x-{\cot }^{2}x\end{array}\right|=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+{\cot }^{2}x\cdot \mathrm{cosec}x\\ {\cos }^{2}x& {\cos }^{2}x& {\mathrm{cosec}}^{2}x\\ 0& 0& -\left({\cot }^{2}x{\cos }^{2}x\right)-\sec x\end{array}\right|$
$f(x)=-\left({\cot }^{2}x{\cos }^{2}x+\sec x\right)\left(\mathrm{cosec}x-{\cos }^{3}x\right)=-\cos x{\sec }^{2}x\left({\cot }^{2}x{\cos }^{2}x+\sec x\right)$
$=-\left({\cos }^{5}x+{\sin }^{2}x\right)=\underset{-\pi }{\overset{\pi }{\int }}\left({\cos }^{5}x+{\sin }^{2}x\right)dx=0-2\underset{0}{\overset{\pi }{\int }}{\sin }^{2}xdx=-4\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}xdx=-\pi$