The area of the region bounded by the curve y =1/(1+(tanx)^1/2) and t…

The given curve is $y =\frac{1}{1+(tanx)^{1/2}}$

$⇒ y =\frac{(cosx)^{1/2}}{(sinx)^{1/2}+(cosx)^{1/2}}$

Area bounded between the intervals x = π/6

and x = π/3 is given by

$A =\int\limits_{π/6}^{π/3}\frac{(cosx)^{1/2}}{(sinx)^{1/2}+(cosx)^{1/2}} dx$

$=\int\limits_{π/6}^{π/3}\frac{(cos(\frac{\pi}{3}+\frac{\pi}{6}-x))^{1/2}}{(sin(\frac{\pi}{3}+\frac{\pi}{6}-x))^{1/2}+(cos(\frac{\pi}{3}+\frac{\pi}{6}-x))^{1/2}}dx$

$=\int\limits_{π/6}^{π/3}\frac{(sinx)^{1/2}}{(sinx)^{1/2}+(cosx)^{1/2}} dx$

Adding $2A =\int\limits_{π/6}^{π/3}dx$

$⇒A = \frac{1}{2}[\frac{\pi}{3}-\frac{\pi}{6}]=\frac{\pi}{12}$ sq. units.

Hence (D) is the correct answer.