A mere word will reveal the solution

$\int _{ \frac { 121\pi }{ 6 } }^{ \frac { 61\pi }{ 3 } }{ (cos{ (x) }+sin({ x) })dx } \\ \int _{ 20\pi +\frac { \pi }{ 6 } }^{ 20\pi +\frac { \pi }{ 3 } }{ (cos(x)+sin(x))dx } \\ \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ cos(x) } dx+\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ sin(x)dx } \\ Integrating\quad and\quad rearranging\quad the\quad terms,\\ sin(\frac { \pi }{ 3 } )-sin(\frac { \pi }{ 6 } )-cos(\frac { \pi }{ 3 } )+cos(\frac { \pi }{ 6 } )\\ \frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ 2 } -\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } \\ \sqrt { 3 } -1\\ Now,\quad we\quad need\quad to\quad bring\quad it\quad in\quad terms\quad of\quad \frac { a }{ 1+\sqrt { b } } \\ Multiplying\quad and\quad dividing\quad \sqrt { 3 } +1,\\ \frac { (\sqrt { 3 } -1)(\sqrt { 3 } +1) }{ (\sqrt { 3 } +1) } \\ \frac { 2 }{ (1+\sqrt { 3 } ) } \\ Comparing,\quad a=2,\quad b=3.\\ So,\quad a+b=5$

CHEERS!!:)

We know that $\displaystyle \cos x + \sin x$ is a periodic function with a period of $\displaystyle 2\pi$ hence the above integral can be written as
$\displaystyle = \int_{20 \pi + \frac{\pi}{6}}^{ 20 \pi + \frac{ \pi}{3}} (\cos x + \sin x) dx$ $\displaystyle = \int_{\frac{\pi}{6}}^{\frac{ \pi}{3}} (\cos x + \sin x) dx$ After solving this we get $\displaystyle \frac{2}{1+ \sqrt{3}}$ so $\displaystyle 2+3=\boxed{5}$

Firstly note that in trigonometry $\frac{61\pi}{3}\equiv\frac{\pi}{3} \ and \ \frac{121\pi}{6}\equiv\frac{\pi}{6}$ This yields $\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (cosx + sinx)dx = [cosx - sinx]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}$ $=|cos \frac{1}{3} \pi + sin \frac{1}{3} \pi| - |cos\frac{1}{6}\pi + sin\frac{1}{6}\pi|$ $= | \frac{1}{2} - \frac{\sqrt3}{2}| - |\frac{\sqrt{3}}{2} - \frac{1}{2}| = \sqrt{3} -1 = \frac{2}{1+\sqrt{3}}$ $\therefore\ a+b = \boxed{5}$