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

We need to evaluate the definite integral: \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} \]
Step 1: Use the property of definite integrals.
We know the property: \[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \] Let \( I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} \).
Using the property with \(a = \frac{\pi}{2}\): \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan\left(\frac{\pi}{2} - x\right)}} \]
Step 2: Simplify \(\tan\left(\frac{\pi}{2} - x\right)\).
We know that \(\tan\left(\frac{\pi}{2} - x\right) = \cot x\).
Therefore: \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\cot x}} \]
Step 3: Express \(\sqrt{\cot x}\) in terms of \(\tan x\).
Since \(\cot x = \frac{1}{\tan x}\), we have \(\sqrt{\cot x} = \frac{1}{\sqrt{\tan x}}\).
Thus: \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \frac{1}{\sqrt{\tan x}}} \]
Step 4: Simplify the integrand.
\[ \frac{1}{1 + \frac{1}{\sqrt{\tan x}}} = \frac{1}{\frac{\sqrt{\tan x} + 1}{\sqrt{\tan x}}} = \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \] Therefore: \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx \]
Step 5: Add the two expressions for \(I\).
We have: \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} \] and \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx \] Adding these two equations: \[ 2I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} + \int_{0}^{\pi/2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx \]
Step 6: Combine the integrals.
\[ 2I = \int_{0}^{\pi/2} \left[ \frac{1}{1 + \sqrt{\tan x}} + \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \right] dx \] The denominators are the same, so: \[ \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} = 1 \] Thus: \[ 2I = \int_{0}^{\pi/2} 1 \, dx \]
Step 7: Evaluate the integral.
\[ \int_{0}^{\pi/2} 1 \, dx = [x]_{0}^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Therefore: \[ 2I = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \]
Step 8: Conclusion.
The value of the given integral is \(\frac{\pi}{4}\). Final Answer: (C) \(\frac{\pi}{4}\)