Question

Let $f(x)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x{\mathrm{cosec}}^{2}x\\ {\cos }^{2}x& {\cos }^{2}x& {\mathrm{cosec}}^{2}x\\ 1& {\cos }^{2}x& {\mathrm{cosec}}^{2}x\end{array}\right|$ then value of $\underset{\pi /4}{\overset{\pi /2}{\int }}f(x)dx$ is

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

Answer 1

Answer: (D)

Applying $R_2\to R_2-R_3$, we get
$f(x)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x{\mathrm{cosec}}^{2}x\\ -{\sin }^{2}x& 0& 0\\ 1& {\cos }^{2}x& {\mathrm{cosec}}^{2}x\end{array}\right|$
$=-\left(-{\sin }^{2}x\right)\left|\begin{array}{cc}\cos x& {\sec }^{2}x+\cot x{\mathrm{cosec}}^{2}x\\ {\cos }^{2}x& {\mathrm{cosec}}^{2}x<br/>\end{array}\right|$
$={\sin }^{2}x[\cos x\mathrm{cosec}x-{\cos }^{2}x\left({\sec }^{2}x+\cot x{\mathrm{cosec}}^{2}x\right)]$
$=\cos x-{\sin }^{2}x-\frac{{\cos }^{3}x}{\sin x}$
$=\cos x-\frac{1}{2}(1-\cos 2x)-\left(\frac{1}{\sin x}-\sin x\right)\cos x$
Thus, $\underset{\pi /4}{\overset{\pi /2}{\int }}f(x)dx=\underset{\pi /4}{\overset{\pi /2}{\int }}\cos xdx-\frac{1}{2}\underset{\pi /4}{\overset{\pi /2}{\int }}dx$
$+\frac{1}{2}\underset{\pi /4}{\overset{\pi /2}{\int }}\cos 2xdx-\underset{\pi /4}{\overset{\pi /2}{\int }}\left(\frac{1}{\sin x}-\sin x\right)\cos xdx$
${=1-\frac{1}{\sqrt{2}}-\frac{1}{2}\left(\frac{\pi }{2}-\frac{\pi }{4}\right)+\frac{1}{4}(0-1)-\left(\log |t|-\frac{t^2}{2}\right)]}_{1/\sqrt{2}}^1$
$=1-\frac{1}{\sqrt{2}}-\frac{\pi }{8}-\frac{1}{2}\log 2$
$\text{ where }t=\sin x$