Integration Grandmaster - P6

If we put $u \; =\; \frac{2\cos\frac12x + x\sin\frac12x}{2\sin\frac12x - x\cos\frac12x}$ then $\begin{aligned} \frac{1}{u}\,\frac{du}{dx} &= \; \frac{\frac12x\cos\frac12x}{2\cos\frac12x + x\sin\frac12x} - \frac{\frac12x\sin\frac12x}{2\sin\frac12x - x\cos\frac12x} \\ & = \; \frac{x\left[\cos\frac12x(2\sin\frac12x - x\cos\frac12x) - \sin\frac12x(2\cos\frac12x + x\sin\frac12x)\right]}{2(2\cos\frac12x + x\sin\frac12x)(2\sin\frac12x - x\cos\frac12x)} \\ & = \; \frac{x^2}{(x^2 - 4)\sin x + 4x\cos x} \end{aligned}$ so that, if we put $\alpha \; = \; \frac{2\tan\frac12}{2\tan\frac12-1}$ then $A \; =\; \int_\alpha^{\frac12\pi}\frac{du}{u} \; = \; \ln\left(\frac{\pi}{2\alpha}\right)$ which makes $\lfloor 10000 A \rfloor = \boxed{-28625}$ .