Question

Find the value of \[ \int_0^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}}. \]

Hint:

Use symmetry and substitution \(x \to \frac{\pi}{2} - x\) in definite integrals involving trigonometric functions.

Answer 1

Let \[ I = \int_0^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}}. \] Make the substitution \( x = \frac{\pi}{2} - t \), so \( dx = -dt \). Note that \[ \tan\left( \frac{\pi}{2} - t \right) = \cot t = \frac{1}{\tan t}. \] Thus, \[ I = \int_{\pi/2}^0 \frac{-dt}{1 + \sqrt{\cot t}} = \int_0^{\pi/2} \frac{dt}{1 + \frac{1}{\sqrt{\tan t}}} = \int_0^{\pi/2} \frac{dt}{1 + \frac{1}{\sqrt{\tan t}}}. \] Simplify the denominator: \[ 1 + \frac{1}{\sqrt{\tan t}} = \frac{\sqrt{\tan t} + 1}{\sqrt{\tan t}}. \] So the integrand becomes \[ \frac{1}{\frac{\sqrt{\tan t} + 1}{\sqrt{\tan t}}} = \frac{\sqrt{\tan t}}{\sqrt{\tan t} + 1}. \] Therefore, \[ I = \int_0^{\pi/2} \frac{\sqrt{\tan t}}{\sqrt{\tan t} + 1} dt. \] Add the two expressions for \( I \): \[ 2I = \int_0^{\pi/2} \left( \frac{1}{1 + \sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \right) dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}. \] Hence, \[ I = \frac{\pi}{4}. \]
Final answer: \[ \boxed{ \frac{\pi}{4}. } \]