Trigo-Fractional Integral

Note that $f(x) = 100 \{\tan x\} \{\cot x\}$ is symmetric over $[0,\tfrac12\pi]$ , and hence $\begin{aligned} I & = \int_0^{\frac12\pi}f(x)\,dx \; = \; 2\int_{\frac14\pi}^{\frac12\pi} f(x)\,dx \; = \; 200\int_{\frac14\pi}^{\frac12\pi} \{\tan x\} \cot x\,dx \\ & = 200\int_{\frac14\pi}^{\frac12\pi} \big(\tan x - \lfloor \tan x \rfloor \big) \cot x\,dx \; = \; 50\pi - 200\int_{\frac14\pi}^{\frac12\pi} \lfloor \tan x \rfloor \cot x\,dx \\ & = 50\pi - 200 \sum_{n=1}^\infty n \int_{\tan^{-1}n}^{\tan^{-1}(n+1)} \cot x\,dx \; = \; 50\pi - 200 \sum_{n=1}^\infty n \Big[\ln(\sin x) \Big]_{\tan^{-1}n}^{\tan^{-1}(n+1)} \\ & = 50\pi - 200 \lim_{N \to \infty}\sum_{n=1}^N n \left[ \ln\left(\frac{n+1}{\sqrt{1+(n+1)^2}}\right) - \ln\left(\frac{n}{\sqrt{1+n^2}}\right)\right] \\ & = 50\pi - 200\lim_{N \to \infty} \left\{ \sum_{n=2}^{N+1} (n-1)\ln\left(\frac{n}{\sqrt{1+n^2}}\right) - \sum_{n=1}^N n \ln\left(\frac{n}{\sqrt{1+n^2}}\right)\right\} \\ & = 50\pi - 200\lim_{N \to \infty}\left\{ N\ln\left(\frac{N+1}{\sqrt{1 + (N+1)^2}}\right) - \sum_{n=1}^N \ln\left(\frac{1}{\sqrt{1+n^{-2}}}\right)\right\} \\ & = 50\pi + 100 \lim_{N \to \infty} N \ln\left(1 + (N+1)^{-2}\right) - 100\lim_{N \to \infty}\ln\left(\prod_{n=1}^N\left(1 + n^{-2}\right)\right) \\ & = 50\pi - 100\ln\left(\prod_{n=1}^\infty \left(1 + n^{-2}\right)\right) \; =\; 50\pi - 100\ln\left(\frac{\sinh \pi}{\pi}\right) \; = \;26.89499282... \end{aligned}$ making the answer $\boxed{27}$ .