Tangle

$\begin{aligned} I & = \int_0^\frac \pi 4 \left(\frac {1-\tan x}{1+\tan x}\right)^4 dx & \small \blue{\text{By reflections }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \int_0^\frac \pi 4 \left(\frac {1-\tan \left(\frac \pi 4 -x\right)}{1+\tan \left(\frac \pi 4 -x\right)}\right)^4 dx \\ & = \int_0^\frac \pi 4 \left(\frac {1-\frac {1-\tan x}{1+\tan x}}{1+\frac {1-\tan x}{1+\tan x}}\right)^4 dx \\ & = \int_0^\frac \pi 4 \tan^4 x \ dx \\ & = \int_0^\frac \pi 4 \tan^2 x (\sec^2 x -1) \ dx \\ & = \int_0^\frac \pi 4 \blue{\tan^2 x} \ d \blue{\tan x} - \int_0^\frac \pi 4 \tan^2 x \ dx \\ & = \frac {\tan^3 x}3 \bigg|_0^\frac \pi 4 - \int_0^\frac \pi 4 (\sec^2 x - 1) \ dx \\ & = \frac 13 - \tan x + x \ \bigg|_0^\frac \pi 4 \\ & = \frac \pi 4 - \frac 23 \approx \boxed{0.119} \end{aligned}$

Reference: Integration tricks -- reflections

Let $\dfrac{π}{4}-x=y$ . Then $dx=-dy$ . So the integral reduces to $\displaystyle \int_0^{\frac{π}{4}} \tan^4 ydy=\displaystyle \int_0^{\frac{π}{4}} \tan^2 y(\sec^2 y-1)dy=\dfrac{π}{4}-\dfrac{2}{3}\approx \boxed {1.1187314967307}$ .