Question

Let $ f(x)=\begin{vmatrix}\sec x & \cos x & \sec ^2 x+\cot x \operatorname{cosec}^2 x \\ \cos ^2 x & \cos ^2 x & \operatorname{cosec}^2 x \\ 1 & \cos ^2 x & \operatorname{cosec}^2 x\end{vmatrix}$ then value of $\int\limits_{\pi / 4}^{\pi / 2} f(x) d x$ is

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

Answer 1

Answer: (D)

Applying $R_2 \rightarrow R_2-R_3$, we get
$f(x)=\begin{vmatrix}\sec x & \cos x & \sec ^2 x+\cot x \operatorname{cosec}^2 x \\-\sin ^2 x & 0 & 0 \\1 & \cos ^2 x & \operatorname{cosec}^2 x\end{vmatrix}$
$=-\left(-\sin ^2 x\right)\begin{vmatrix}\cos x & \sec ^2 x+\cot x \operatorname{cosec}^2 x \\\cos ^2 x & \operatorname{cosec}^2 x
\end{vmatrix}$
$= \sin ^2 x[\cos x \operatorname{cosec} x \left.-\cos ^2 x\left(\sec ^2 x+\cot x \operatorname{cosec}^2 x\right)\right]$
$=\cos x-\sin ^2 x-\frac{\cos ^3 x}{\sin x} $
$=\cos x-\frac{1}{2}(1-\cos 2 x)-\left(\frac{1}{\sin x}-\sin x\right) \cos x $
Thus, $\int\limits_{\pi / 4}^{\pi / 2} f(x) d x=\int\limits_{\pi / 4}^{\pi / 2} \cos x d x-\frac{1}{2} \int\limits_{\pi / 4}^{\pi / 2} d x$
$+\frac{1}{2} \int\limits_{\pi / 4}^{\pi / 2} \cos 2 x d x-\int\limits_{\pi / 4}^{\pi / 2}\left(\frac{1}{\sin x}-\sin x\right) \cos x d x $
$\left.=1-\frac{1}{\sqrt{2}}-\frac{1}{2}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)+\frac{1}{4}(0-1)-\left(\log |t|-\frac{t^2}{2}\right)\right]_{1 / \sqrt{2}}^1$
$=1-\frac{1}{\sqrt{2}}-\frac{\pi}{8}-\frac{1}{2} \log 2 $
$ \text { where } t=\sin x$