What exactly is it?

@Calvin Lin and @Eilon Lavi here is the solution,

$\begin{aligned} f(x) &=& x - [x], \text { if } f[x] \text{ is odd} \\ f(x) &=& 1 + [x] - x, \text { if } f[x] \text{ is even} \\ \end{aligned}$

$\Rightarrow f\left( x \right) =\left\{ \begin{matrix} x-1,\quad \quad 1\le x<2 \\ 1-x,\quad \quad 0\le x<1 \end{matrix} \right\}$

$f(x) \ is \ periodic \ with \ period \ 2$

$\therefore \quad I=\int _{ -10 }^{ 10 }{ f(x).cos\quad \pi x\quad dx }$

$I=2\int _{ 0 }^{ 10 }{ f(x).cos\quad \pi x } \quad dx\quad =\quad 2\times 5\int _{ 0 }^{ 2 }{ f(x).cos\quad \pi x\quad } dx$

$=10\left[ \int _{ 0 }^{ 1 }{ (1-x).cos\quad \pi x } dx \ + \ \int _{ 1 }^{ 2 }{ (x-1).cos\quad \pi x } dx \right] =10({ I }_{ 1 }{ +I }_{ 2 })$

$Now,{ \quad I }_{ 2 }=\int _{ 1 }^{ 2 }{ (x-1)cos\quad \pi x\quad dx } \quad \quad \quad \quad \quad put\quad x-1=t$

${ I }_{ 2 }=-\int _{ 0 }^{ 1 }{ t.cos\quad \pi t\quad dt } =-\int _{ 0 }^{ 1 }{ x.cos\quad \pi x\quad dx }$

$Also,{ \quad I }_{ 1 }=\int _{ 0 }^{ 1 }{ (1-x)cos\quad \pi x\quad dx } =-\int _{ 0 }^{ 1 }{ x.cos\quad \pi x\quad dx }$

$\Rightarrow I=10 \left[ -2\int _{ 0 }^{ 1 }{ x.cos\quad \pi x\quad dx } \right]$

$I=-20{ \left[ x\frac { sin\quad \pi x }{ \pi } +\frac { cos\quad \pi x }{ { \pi }^{ 2 } } \right] }_{ 0 }^{ 1 }$

$I=-20{ \left[ -\frac { 1 }{ { \pi }^{ 2 } } -\frac { 1 }{ { \pi }^{ 2 } } \right] }$

$\Rightarrow { \frac { { \pi }^{ 2 } }{ 10 } I\quad =\quad \boxed{4} }$