How will you Integrate it?

$\begin{aligned} \left(1-f(x)\right)f(x+\pi) & = 1 + f(x) \\ f(x+\pi) & = \frac{1+f(x)}{1-f(x)} \\ \Rightarrow f([x+\pi] + \pi) & = \frac{1+f(x+\pi)}{1-f(x+\pi)} \\ f(x+2\pi) & = \frac{1+ \frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}} = \frac{1-f(x)+1+f(x)}{1-f(x)-1-f(x)} = \frac{2}{-2f(x)} \\ \Rightarrow f(x+2\pi) & = - \frac{1}{f(x)} \\ f(x) & = - \frac{1}{f(x+2\pi)} \\ \\ \Rightarrow f(x + 5\pi) & = - \frac{1}{f(x+ 3 \pi)} = f(x + \pi) \\ \Rightarrow f(x) & = f(x+\color{#3D99F6}{4\pi}) \quad \color{#3D99F6}{\Rightarrow f(x) \text{ is a periodic function with a period of } 4 \pi.} \end{aligned}$

Since $\displaystyle \int_\pi^{5\pi} f(x) \space dx = 2$ , integral of $f(x)$ over every $4\pi$ has a value of $2$ .

Therefore, $\displaystyle \int_{3\pi}^{2015 \pi} f(x) \space dx = \frac{2015-3}{4} \times 2 = \boxed{1006}$