Answer
$$\frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\sqrt {\tan x} {{\sec }^2}x} dx \cr
& u = \tan x,{\text{ then }}du = {\sec ^2}xdx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \pi /4,{\text{ }}u = \tan \left( {\pi /4} \right) = 1 \cr
& {\text{if }}x = 0,{\text{ }}u = \tan \left( 0 \right) = 0 \cr
& {\text{so}} \cr
& \int_0^{\pi /4} {\sqrt {\tan x} {{\sec }^2}x} dx = \int_0^1 {\sqrt u } du \cr
& = \int_0^1 {{u^{1/2}}du} \cr
& {\text{find the antiderivative }} \cr
& = \left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_0^1 \cr
& = \frac{2}{3}\left( {{u^{3/2}}} \right)_0^1 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{2}{3}\left( {{{\left( 1 \right)}^{3/2}} - {{\left( 0 \right)}^{3/2}}} \right) \cr
& = \frac{2}{3}\left( 1 \right) \cr
& = \frac{2}{3} \cr} $$
