`int_(pi/6)^(pi/3) (sin x+ cos x)/(sqrt(sin 2x))dx`

Explanation

`int_(pi//6)^(pi//3) (sinx +cos x)/(sqrt(sin2x)) dx `
we know that
`(sin x-cos x)^(2) =sin^(2) x+cos^(2) x-2 sinx cos x`
`rArr (sin x-cos x)^(2) =1- sin 2x`
`rArr" "sin 2x =1 -(sin x-cos x)^(2)`
`"Let " sin x=cos x=t`
`rArr (cos x+sin x) dx=dt`
`" and " x=(pi)/(6) rArr t=sin.(pi)/(6) -cos .(pi)/(6) =(1)/(2) -(sqrt(3))/(2)`
`" and " x=(pi)/(3) rArr t=sin .(pi)/(3) -cos .(pi)/(3)=(sqrt(3))/(2)-(1)/(2)`
`=int_((1-sqrt(3))//2)^((sqrt(3)-1)//2).(dt)/(sqrt(1-t^(2)))`
`=[sin^(-1) t] _((1-sqrt(3))//2)^((sqrt(3)-1)//2)`
`=[sin^(-1) ((sqrt(3)-1)/(2))-sin^(-1) {(-(sqrt(3)-1)/(2))}]`
`=[sin^(-1) ((sqrt(3)-1)/(2))+sin^(-1) ((sqrt(3)-1)/(2))]`
`[ :' sin^(-1) (-0) =-sin^(-1)0]`
`=2sin^(-1) ((sqrt(3)-1)/(2))`