The area of the region bounded by the curve y = \(\frac{1}{1 + \sqrt{\tan {x}}}\) and the x-axis between the ordinates x = \(\frac{\pi}{6}\) and x = \(\frac{\pi}{3}\) is : 

None of these

y = \(\frac{1}{1 + \sqrt{\tan {x}}}\) 

  = \(\frac{\sqrt{\cos {x}}}{\sqrt{\ sin {x}} + \sqrt{\cos {x}}}\)

Area bounded between the intervals x = \(\frac{\pi}{6}\) and x = \(\frac{\pi}{3}\) is given by 

A = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x [\(\frac{\sqrt{\cos {x}}}{\sqrt{\ sin {x}} + \sqrt{\cos {x}}}\)]

   = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x [\(\frac{\sqrt{\cos {\frac{\pi}{3} + \frac{\pi}{6} -x}}}{\sqrt{\ sin {\frac{\pi}{3} + \frac{\pi}{6} -x}} + \sqrt{\cos {\frac{\pi}{3} + \frac{\pi}{6} -x}}}\)]

   = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x [\(\frac{\sqrt{\sin {x}}}{\sqrt{\ sin {x}} + \sqrt{\cos {x}}}\)]

2A = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}d\)x 

⇒ A = \(\frac{1}{2} [\frac{\pi}{3} - \frac{\pi}{6}] \)

      = \(\frac{\pi}{12}\) sq. units.