Correct Answer - A
Let ` l = 2/3 int _(0)^(pi//2) (sqrt(cos theta))/((sqrt(sin theta)+sqrt(cos theta))) d theta " "` …(i)
and ` l = 2/3 int _(0)^(pi//2) (sqrt(cos (pi//2-theta)))/(sqrt(sin (pi//2-theta)) + sqrt(cos (pi//2- theta))) d theta `
` rArr l = 2/3 int _(0)^(pi//2) (sqrt(sintheta))/(sqrt(sin theta )+sqrt(cos theta)) d theta " "` (ii)
On adding Eqs. (i) and (ii) , we get
` 2l = 2/3 int _(0)^(pi//2) 1 d theta = 2/3 [ theta ] _(0)^(pi//2)`
` rArr l = 1/3 . pi/2 = pi/6`