Answer
$$\frac{{\sqrt 2 - 1}}{\pi }$$
Work Step by Step
$$\eqalign{
& \int_0^{1/4} {\sec \pi x\tan \pi x} dx \cr
& {\text{Write the integrand as}} \cr
& = \frac{1}{\pi }\int_0^{1/4} {\left( {\sec \pi x} \right)\left( {\tan \pi x} \right)} \left( \pi \right)dx \cr
& {\text{Using the formula }}\int {\sec u\tan u} du = \sec u + C \cr
& = \frac{1}{\pi }\left( {\sec \pi x} \right)_0^{1/4} \cr
& {\text{Evaluate and simplify}} \cr
& = \frac{1}{\pi }\left( {\sec \frac{\pi }{4} - \sec 0} \right) \cr
& = \frac{1}{\pi }\left( {\sqrt 2 - 1} \right) \cr
& = \frac{{\sqrt 2 - 1}}{\pi } \cr} $$
