integral evaluation

We have $I=\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ 1+\sqrt { tan(x) } } }$ (i)

Using the property $\int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx }$ we have :

$I=\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ 1+\sqrt { tan(\frac { \pi }{ 2 } -x) } } } =\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ 1+\frac { 1 }{ \sqrt { tan(x) } } } } =\int _{ \pi /6 }^{ \pi /3 }{ \frac { \sqrt { tan(x) } dx }{ 1+\sqrt { tan(x) } } }$ (ii)

Adding (i) and (ii) we get

$2I=\int _{ \pi /6 }^{ \pi /2 }{ dx } =\frac { \pi }{ 6 }$

$\Rightarrow \boxed{I=\frac{\pi}{12}}$