Explanation

(a)
Let $I = \int_{\pi/3}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}}$
$\int_{\pi/6}^{\pi/3}{\left( \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right)dx\ \ ...(i)}$
$\Rightarrow I = \int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}$ ...(ii)
$\left( \because\int_{a}^{b}{f(x)dx = \int_{a}^{b}{f(a + b - x)dx}} \right)$
On adding Eqs. (i) and (ii), we get
$2I = \int_{\pi/6}^{\pi/3}{1\ dx = \lbrack x\rbrack_{\pi/6}^{\pi/3} = \frac{\pi}{6}}$
$\Rightarrow \ I = \frac{\pi}{12}$