Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /4}^{\pi /4} {{{\sec }^2}x} dx \cr
& {\text{Identify the symmetry in the function }}{\sec ^2}x,{\text{ evaluate }}f\left( { - x} \right) \cr
& f\left( { - x} \right) = {\sec ^2}\left( { - x} \right) \cr
& f\left( { - x} \right) = \frac{1}{{{{\left[ {\cos \left( { - x} \right)} \right]}^2}}} \cr
& f\left( { - x} \right) = \frac{1}{{{{\left[ {\cos \left( x \right)} \right]}^2}}} \cr
& f\left( { - x} \right) = \frac{1}{{{{\cos }^2}x}} \cr
& f\left( { - x} \right) = {\sec ^2}x \cr
& f\left( { - x} \right) = f\left( x \right) \cr
& {\text{Therefore }}{\sec ^2}x{\text{ is an even function}}{\text{.}} \cr
& {\text{Use the integral property }}\int_{ - a}^a {f\left( x \right)} dx = 2\int_0^2 {f\left( x \right)} dx,{\text{ }}f{\text{ is even}} \cr
& \int_{ - \pi /4}^{\pi /4} {{{\sec }^2}x} dx = 2\int_0^{\pi /4} {{{\sec }^2}x} dx \cr
& {\text{Integrating}} \cr
& = 2\left[ {\tan x} \right]_0^{\pi /4} \cr
& = 2\left[ {\tan \left( {\frac{\pi }{4}} \right) - \tan \left( 0 \right)} \right] \cr
& = 2\left( {1 - 0} \right) \cr
& = 2 \cr} $$
