Discussion Forum

 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

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

Answer:

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