Answer:
Option D
Explanation:
Let I= $\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{tan x}}$ .........(i)
$\therefore$ I =$\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{tan (\frac{\pi}{2}-x)}}$
= $\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{cot x}}$
$\Rightarrow$ $ I= \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}}$........(ii)
On adding Eqs.(i) and (ii) , we get
$2 I= \int_{\pi/6}^{\pi/3} dx\Rightarrow2I=[x]^{\pi/3}_{\pi/6}$
$\Rightarrow I=\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{6}\right]=\frac{\pi}{12}$
Statement I false
But $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$
true statement by property of definite integrate
