<s>[INST] Find the integral part of the product $\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdots \frac{2016}{2015}.$ 
Find the integral part of the following number $$T = \dfrac{2}{1} \cdot \dfrac{4}{3} \cdot \dfrac{6}{5} \cdot \dfrac{8}{7} \cdots \dfrac{2016}{2015}.$$

We can show that $T = 2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$, since $$\int_0^{\frac{\pi}{2}} \sin^{2n+1}(x) dx = \dfrac{2}{3} \cdot \dfrac{4}{5} \cdot \dfrac{6}{7} \cdots \dfrac{2n}{2n+1},$$ but how do we calculate the integral part of $2017\int_{0}^{\frac{\pi}{2}} \sin^{2017}(x)dx$?
 [/INST] Although @winther and @achillehui did this the smart way and used the standard approximations I took the "battering ram" approach by first rewriting the product in the form
\begin{equation}
P=\dfrac{2^{2016}(1008!)^2}{2016!}
\end{equation}
This gave
\begin{equation}
\ln P=2016\ln(2)+\sum_{k=1}^{1008}\ln\left(\dfrac{k}{k+1008}\right)\approx4.030351009
\end{equation}
which gives $P\approx56.28066279$
Rather than write a short program to calculate the sum I used a spreadsheet but still obtained a result in agreement with approximation formulas.
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