Question

Let $f(x)=\begin{vmatrix}\sec x & \cos x & \sec ^2 x+\cot x \cdot \csc x \\ \cos ^2 x & \cos ^2 x & \csc ^2 x \\ 1 & \cos ^2 x & \cos ^2 x\end{vmatrix}$ then $\int_{-\pi}^\pi f(x) d x$ has the value equal to

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

Answer 1

Answer: (C)

$R _3 \rightarrow R _3-(\cos x ) R _1$
$f(x)=\begin{vmatrix}\sec x & \cos x & \left(\sec ^2 x+\cot x \cdot \operatorname{cosec} x\right) \\ \cos ^2 x & \cos ^2 x & \operatorname{cosec} 2 \\ 0 & 0 & \cos ^2 x-\sec x-\cot ^2 x\end{vmatrix}=\begin{vmatrix}\sec x & \cos x & \sec ^2 x+\cot ^2 x \cdot \operatorname{cosec} x \\ \cos ^2 x & \cos ^2 x & \operatorname{cosec}^2 x \\ 0 & 0 & -\left(\cot ^2 x \cos ^2 x\right)-\sec x\end{vmatrix}$
$f(x)=-\left(\cot ^2 x \cos ^2 x+\sec x\right)\left(\operatorname{cosec} x-\cos ^3 x\right)=-\cos x \sec ^2 x\left(\cot ^2 x \cos ^2 x+\sec x\right)$
$=-\left(\cos ^5 x+\sin ^2 x\right)=\int\limits_{-\pi}^\pi\left(\cos ^5 x+\sin ^2 x\right) d x=0-2 \int\limits_0^\pi \sin ^2 x d x=-4 \int\limits_0^{\pi / 2} \sin ^2 x d x=-\pi$