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The value of the integral

Statement I

$\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}}$ is equal to $ \pi/6$

Statement II

$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x)dx$

A) Statement I is true, statement II is true, statement II is a correct explanation for statement I

B) Statement I is true, statement II is true, statement II is not a correct explanation for statement I

C) Statement I is true, statement II is false

D) Statement I is false, statement II is true

Option D

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

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