Definite integral Problem-1

$\int _{ 0 }^{ 2\pi }{ \frac { dx }{ 31+5cosx+6sinx } } \\ \int _{ 0 }^{ 2\pi }{ \frac { dx }{ 31+5\frac { 1-{ tan }^{ 2 }\frac { x }{ 2 } }{ 1+{ tan }^{ 2 }\frac { x }{ 2 } } +6\frac { 2{ ta }n\frac { x }{ 2 } }{ 1+{ tan }^{ 2 }\frac { x }{ 2 } } } } \\ \int _{ 0 }^{ 2\pi }{ \frac { { sec }^{ 2 }\frac { x }{ 2 } .dx }{ 26{ tan }^{ 2 }\frac { x }{ 2 } +12tan\frac { x }{ 2 } +36 } } \\ now\quad put\quad tan\frac { x }{ 2 } =t\\ { sec }^{ 2 }\frac { x }{ 2 } .dx=2.dt\\ \int { \frac { dt }{ 13{ t }^{ 2 }+6t+18 } } \\ \frac { 1 }{ 13 } \int { \frac { dt }{ { (t+\frac { 3 }{ 13 } ) }^{ 2 }+{ (\frac { 15 }{ 13 } ) }^{ 2 } } } \\ \frac { 1 }{ 15 } { tan }^{ -1 }(\frac { 13tan\frac { x }{ 2 } +3 }{ 15 } )\overset { 2\pi }{ \underset { 0 }{ | } } \\ =\quad \frac { \Pi }{ 15 } \\ NOTE:\\ If\quad you\quad directly\quad put\quad the\quad limits,\\ then\quad you\quad will\quad get\quad answer\quad =\quad 0\\ So\quad rather\quad than\quad putting\quad the\quad limits\quad directly,\\ draw\quad the\quad graph\quad of\quad this\quad function\quad to\\ identify\quad the\quad range\quad over\quad which\quad function\\ is\quad monotonous.\\$