MCQ
Let $f(x) = \left| {\begin{array}{*{20}{c}}{\,\sec x}&{\cos x}&{{{\sec }^2}x + \cot x\,{\rm{cosec}}\,x\,}\\{{{\cos }^2}x}&{{{\cos }^2}x}&{{\rm{cose}}{{\rm{c}}^2}x}\\1&{{{\cos }^2}x}&{{{\cos }^2}x}\end{array}} \right|\,,$ then $\int_0^{\pi /2} {\,f(x)\,dx = } $
  • A
    $\frac{\pi }{4} + \frac{8}{{15}}$
  • B
    $\frac{\pi }{4} - \frac{8}{{15}}$
  • $ - \frac{\pi }{4} - \frac{8}{{15}}$
  • D
    $ - \frac{\pi }{4} + \frac{8}{{15}}$

Answer

Correct option: C.
$ - \frac{\pi }{4} - \frac{8}{{15}}$
c
(c)Applying ${R_1} \to {R_1} - \sec x{R_3},$we get
$f(x) = - {\sin ^2}x - {\cos ^5}x$
Thus $\int_0^{\pi /2} {f(x)dx = - \int_0^{\pi /2} {({{\sin }^2}x + {{\cos }^5}x)dx} } $
$ = - \left[ {\frac{\pi }{4} + \frac{8}{{15}}} \right]$=$\frac{{ - \pi }}{4} - \frac{8}{{15}}$.

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