\(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}=\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\frac{\sqrt{\sin x}}{\sqrt{\cos x}}}=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\)
Let \(I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) .....(1)
By using properly \(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\)
\(I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) .....(2)
\((1) +(2)\)
\(2 I=\int_{\pi / 6}^{\pi / 3} d x \Rightarrow 2 I=[x]_{\pi / 6}^{\pi / 3} \Rightarrow 2 I=\frac{\pi}{3}-\frac{\pi}{6}\)
\(\Rightarrow 2 I=\frac{2 \pi-\pi}{6} \Rightarrow I=\frac{\pi}{12}\)
