If $f\left( x \right) = \,\,\left| {\,\begin{array}{*{20}{c}} {\sec \,\,x}&{\cos \,\,x}&{{{\sec }^2}\,\,x\,\, + \,\,{\rm{cosec}}\,\,x\,\,\cot \,\,x}\\ {{{\cos }^2}\,\,x}&{{{\cos }^2}\,\,x}&{{\rm{cose}}{{\rm{c}}^2}\,\,x}\\ 1&{{{\cos }^2}\,\,x}&{{{\cos }^2}\,\,x} \end{array}\,} \right|$, then $\mathop \smallint \limits_0^{\pi /2} \,\,f\left( x \right)\,\,dx$ equals

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

f\left( x \right) = \,\,\left| {\,\begin{array}{cccccccccccccccccccc} {\sec \,\,x}&{\cos \,\,x}&{{{\sec }^2}\,\,x\,\, + \,\,\mathrm{cosec}\,\,x\,\,\cot \,\,x}\\ {{{\cos }^2}\,\,x}&{{{\cos }^2}\,\,x}&{\mathrm{cose}{\mathrm{c}^2}\,\,x}\\ 1&{{{\cos }^2}\,\,x}&{{{\cos }^2}\,\,x} \end{array}\,} \right|

, then

\int \limits_0^{\pi /2} \,\,f\left( x \right)\,\,dx

equals

View answer and explanation

Correct option: C

- \,\,\left( {\frac{\pi }{4}\,\, + \,\,\frac{8}{{15}}} \right)

Option C (

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

) is correct because after evaluating the determinant and integrating the resulting trigonometric functions, we obtain this value.