Integration Grandmaster - P10

Note first that

$A \; = \; \int_0^1 \frac{\cos 2x - \tan x\,\cot(\tan x)}{\sin 2x - \tan(\tan x)\,\ln(\cos^2x)}\,dx \; = \; \int_0^1 \frac{\cos 2x\cot(\tan x) - \tan x\,\cot^2(\tan x)}{\sin 2x \cot(\tan x) - 2\ln(\cos x)}\,dx$ and putting $u = \sin2x \cot(\tan x) - 2\ln(\cos x)$ for $0 < u < \tan^{-1}\pi$ gives $\lim_{x \to 0}u = 2$ and $\begin{aligned} \frac{du}{dx} & = \; 2\cos2x \cot(\tan x) - \sin2x \mathrm{cosec}^2(\tan x)\,\sec^2x + 2\tan x \; = \; 2\cos 2x \cot(\tan x) - 2\tan x\big(\mathrm{cosec}^2(\tan x) - 1\big)\\ & = \; 2\big[\cos2x \cot(\tan x) - \tan x \cot^2(\tan x)\big] \end{aligned}$ and hence, putting $\alpha = \sin2 \cot(\tan 1) - 2\ln(\cos 1)$ we see that $A \; = \; \int_2^\alpha \frac{du}{2u} \; = \; \tfrac12\ln\big(\tfrac{\alpha}{2}\big) \; = \; \tfrac12\ln\left(\tfrac12\sin 2 \cot(\tan 1) - \ln(\cos 1)\right)$ which makes $\lfloor 10000 A\rfloor = \boxed{-2377}$ .