Answer
$$8$$
Work Step by Step
$$\eqalign{
& \int_{1 - \pi }^{1 + \pi } {{{\sec }^2}\left( {\frac{1}{4}x - \frac{1}{4}} \right)} dx \cr
& u = \frac{1}{4}x - \frac{1}{4},{\text{ then }}du = \frac{1}{4}dx,{\text{ }}4du = xdx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 1 + \pi ,{\text{ }}u = \frac{1}{4}\left( {1 + \pi } \right) - \frac{1}{4} = \frac{\pi }{4} \cr
& {\text{if }}x = 1 - \pi ,{\text{ }}u = \frac{1}{4}\left( {1 - \pi } \right) - \frac{1}{4} = - \frac{\pi }{4} \cr
& {\text{so}} \cr
& \int_{1 - \pi }^{1 + \pi } {{{\sec }^2}\left( {\frac{1}{4}x - \frac{1}{4}} \right)} dx = \int_{ - \pi /4}^{\pi /4} {{{\sec }^2}\left( u \right)} \left( {4du} \right) \cr
& = 4\int_{ - \pi /4}^{\pi /4} {{{\sec }^2}\left( u \right)} du \cr
& {\text{find the antiderivative }} \cr
& = 4\left( {{{\tan }^2}u} \right)_{ - \pi /4}^{\pi /4} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = 4\left( {{{\tan }^2}\left( {\frac{\pi }{4}} \right) - {{\tan }^2}\left( { - \frac{\pi }{4}} \right)} \right) \cr
& = 4\left( {1 - \left( { - 1} \right)} \right) \cr
& = 8 \cr} $$
