Question

The value of \( \displaystyle \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} \) will be

• A. 0
• B. \(\frac{\pi}{2}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{8}\)

Hint:

For integrals of the form \(\int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}}\), use the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\) and add the two expressions to simplify.

Answer 1

The Correct Option is C

The given integral is:
\[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} \] To evaluate this integral, we make a substitution. Let: \[ u = \frac{\pi}{2} - x \quad \text{so that} \quad du = -dx \] When \(x = 0\), \(u = \frac{\pi}{2}\), and when \(x = \frac{\pi}{2}\), \(u = 0\). The integral becomes: \[ I = \int_{\pi/2}^{0} \frac{-du}{1 + \sqrt{\tan\left( \frac{\pi}{2} - u \right)}} \] Using the identity \( \tan\left( \frac{\pi}{2} - u \right) = \cot(u) \), we get: \[ I = \int_{0}^{\pi/2} \frac{du}{1 + \sqrt{\cot u}} \] Now, the original integral is transformed into: \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\cot x}} \] Adding these two expressions for \(I\), we obtain: \[ 2I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} + \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\cot x}} \] Simplifying the sum of the two integrals: \[ 2I = \int_{0}^{\pi/2} \left( \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{\cot x}} \right) dx \] After simplifying the expression inside the integral, we find that: \[ 2I = \int_{0}^{\pi/2} 1 \, dx \] So, \[ 2I = \left[ x \right]_{0}^{\pi/2} = \frac{\pi}{2} \] Therefore, \[ I = \frac{\pi}{4} \] Final Answer:
The value of the integral is \( \boxed{\frac{\pi}{4}} \).