Question

Let$f\left(x\right)=\left|\begin{array}{ccc}cosecx& \sin x& cosec{}^{2}x+\tan x\sec x\\ {\sin }^{2}x& {\sin }^{2}x& {\sec }^{2}x\\ 1& {\sin }^{2}x& {\sin }^{2}x\end{array}\right|$
then ${\int }_0^{\pi /2}f\left(x\right)dx$ equals?

• A. $-(\frac{\pi }{4}+\frac{8}{15})$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{4}+\frac{1}{5}$
• D. None of the above

Answer 1

Answer: (A)

$-(\frac{\pi }{4}+\frac{8}{15})$

$f\left(x\right)=\left|\begin{array}{ccc}cosecx& \sin x& cosec{}^{2}x+\tan x\sec x\\ {\sin }^{2}x& {\sin }^{2}x& {\sec }^{2}x\\ 1& {\sin }^{2}x& {\sin }^{2}x\end{array}\right|$
Applying $C_2\to C_2-{\sin }^{2}x{C}_1$
$f\left(x\right)=\left|\begin{array}{ccc}cosecx& 0& cosec{}^{2}x+\tan x\sec x\\ {\sin }^{2}x& {\sin }^{2}x{\cos }^{2}x& {\sec }^{2}x\\ 1& 0& {\sin }^{2}x\end{array}\right|=-\left({\sin }^{2}x{\cos }^{2}x\right)(\sin x-cosec{}^{2}x-\tan x\sec x)=-{\sin }^{5}x-{\cos }^{2}x$
Therefore on solving
${\int }_0^{\pi /2}f\left(x\right)dx=-(\frac{\pi }{4}+\frac{8}{15})$