For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on the interval $[-10,10]$ by $f(x)=\begin{cases} x-[x] & \text{if } [x] \text{ is odd,}\\ 1+[x]-x & \text{if } [x] \text{ is even} \end{cases}$ Then the value of $\frac{\pi^{2}}{10}\int_{-10}^{10} f(x)\cos \pi x\,dx$ is

$f(x)=x-[x] \text{ if } [x] \text{ odd};\quad 1+[x]-x \text{ if } [x] \text{ even}$

$x \in [-10,10]$

$\text{Find } \dfrac{\pi^{2}}{10}\int_{-10}^{10} f(x)\cos\pi x\,dx$

$f \text{ has period } 2;\ f \text{ and } \cos\pi x \text{ are even}$

$\text{Reduce } \int_{-10}^{10} \text{ down to } \int_{0}^{2}$

$[0,2):\ f(x)=|x-1|$

$f \text{ has period } 2,\qquad f(-x)=f(x)$

$f(x)\cos\pi x \text{ is even}$

$\int_{-10}^{10} f\cos\pi x\,dx = 2\int_{0}^{10} f\cos\pi x\,dx$

$f(x)\cos\pi x \text{ has period } 2$

$\int_{0}^{10} f\cos\pi x\,dx = 5\int_{0}^{2} f\cos\pi x\,dx$

$\int_{0}^{1}(1-x)\cos\pi x\,dx = 0-\left(-\dfrac{2}{\pi^{2}}\right)=\dfrac{2}{\pi^{2}}$

$\int_{1}^{2}(x-1)\cos\pi x\,dx = \dfrac{2}{\pi^{2}}$

$\int_{0}^{2} f\cos\pi x\,dx = \dfrac{4}{\pi^{2}}$

$\int_{-10}^{10} f\cos\pi x\,dx = 2\cdot5\cdot\dfrac{4}{\pi^{2}} = \dfrac{40}{\pi^{2}}$

$\dfrac{\pi^{2}}{10}\cdot\dfrac{40}{\pi^{2}} = 4$

$\boxed{\text{Answer: } 4}$