*Pukes Mathematically I

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {dx}{1+\frac 1{\sqrt[3]{\tan x}}} & \small \color{#3D99F6}{\text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^\frac \pi 2 \frac 1{1+\frac 1{\sqrt[3]{\tan x}}} + \frac 1{1+\frac 1{\sqrt[3]{\tan \left( \frac \pi 2 - x\right)}}} dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac {\sqrt[3]{\tan x}}{\sqrt[3]{\tan x}+1} + \frac 1{1+\frac 1{\sqrt[3]{\cot x}}} dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac {\sqrt[3]{\tan x}}{\sqrt[3]{\tan x}+1} + \frac 1{1+\sqrt[3]{\tan x}} dx \\ & = \frac 12 \int_0^\frac \pi 2 1 \ dx \\ & = \frac \pi 4 \end{aligned}$

$\implies \lfloor 1000I \rfloor = \left \lfloor 1000 \times \frac \pi 4 \right \rfloor = \boxed{785}$

Let $I(a)=\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\tan^{a}(x)}$

Using the fact that $\tan(x)=\frac{\sin(x)}{\cos(x)}$ :

$I(a)=\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\frac{\sin^{a}(x)}{\cos^{a}(x)}}=\int_{0}^{\frac{\pi}{2}}\frac{dx}{\frac{\cos^{a}(x)+\sin^{a}(x)}{\cos^{a}(x)}}=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{a}(x)dx}{\cos^{a}(x)+\sin^{a}(x)}$

Now, let's integrate backwards, letting $x=\frac{\pi}{2}-u$ , and $du=-dx$ .

$I(a)=\int_{\frac{\pi}{2}}^{0}-\frac{\cos^{a}\left(\frac{\pi}{2}-u\right)du}{\cos^{a}\left(\frac{\pi}{2}-u\right)+\sin^{a}\left(\frac{\pi}{2}-u\right)}=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{a}(u)du}{\cos^{a}(u)+\sin^{a}(u)}$

Now adding the first and second expressions for $I(a)$ :

$2I(a)=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{a}(x)dx}{\cos^{a}(x)+\sin^{a}(x)}+\frac{\sin^{a}(x)dx}{\cos^{a}(x)+\sin^{a}(x)}=\int_{0}^{\frac{\pi}{2}}dx=\frac{\pi}{2}$ $I(a)=\frac{\pi}{4}$

Our original integral is simply $I\left(\frac{-1}{3}\right)=\frac{\pi}{4}$

$I=\frac{\pi}{4} \approx .7853$ , so $\lfloor 1000I \rfloor = \boxed{785}$ .