Let I=∫x/3x/6dx1+√tanxI=∫π/3x/6dx1+√tan(π6+π3−x)[Byusing property ∫baf(x)dx=∫baf(a+b−x)dx]…(i)I=∫π/3π/6dx1+√tan(π2−x)I=∫π/3π/6dx1+1√cos2=∫π/3π/6dx1+√cotxI=∫π/3π/6√tanx1+√tanxdx
Adding (i) and (ii), we get
2I=∫π/3π/6(1+√tanx)(1+√tanx)dx=∫π/3π/6dx=[x]π/3x/6=π3−π6=π62I=π6 or I=π12