$\text{ The value of the definite integral, }{\int }_{{\theta }_1}^{{\theta }_2}\frac{\mathrm{d}\theta }{1+\tan \theta }=\frac{501\pi }{\mathrm{K}}$$\text{ where }{\theta }_2=\frac{1003\pi }{2008}\text{ and }{\theta }_1=\frac{\pi }{2008}\text{ . The value of }K\text{ equals }$

$\begin{array}{l}\therefore I={\int }_{{\theta }_1}^{{\theta }_2}\frac{d\theta }{1+\tan \left(\frac{\pi }{2}-\theta \right)}={\int }_{{\theta }_1}^{{\theta }_2}\frac{\tan \theta d\theta }{1+\tan \theta }\\ \text{ and also }I={\int }_{\theta 1}^{\theta 2}\frac{d\theta }{1+\tan \theta }\\ \therefore 2\text{I}={\int }_{{\theta }_1}^{{\theta }_2} \text{d}\theta ={\theta }_2-{\theta }_1=\frac{1002\pi }{2008}\Rightarrow \text{I}=\frac{501\pi }{2008}\\ \text{ Hence, K}=2008\end{array}$