Evaluate `int_((pi)/(6))^((pi)/(3))(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx`

Explanation

We have `I=int_(pi//6)^(pi//3)(sqrt((sinx))dx)/(sqrt((sinx))+sqrt((cosx)))`…………..1
or `I=int_(pi//6)^(pi/3)(sqrt((cosx))dx)/(sqrt((cosx))+sqrt((sinx)))` (Replacing `x` by `(pi)/2-x)`…………..2
Adding 1 and 2 we get
`2I=int_(pi//6)^(pi//3)(sqrt((sinx))+sqrt((cosx)))/(sqrt((cosx))+sqrt((sinx))) dx`
`int_(pi//6)^(pi//3)dx=[x]_(pi//6)^(pi//3)=pi//3-pi//6=pi//6`
Hence `I=pi//12`