Question

If \( I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}} \), then prove that \( I = \frac{\pi}{12} \).

Hint:

Whenever you see a definite integral from \(a\) to \(b\) and the sum \(a+b\) is a convenient value like \( \pi/2 \) or \( \pi \), you should immediately consider using the property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \). This technique is very common for integrals involving trigonometric functions.

Answer 1

Step 1: Understanding the Concept:
This is a problem involving a definite integral that can be solved elegantly using a specific property of definite integrals, often referred to as the "King's Rule".
Step 2: Key Formula or Approach:
We will use the property: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \).
Step 3: Detailed Proof:
Let the given integral be: \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} \, dx \quad \cdots (1) \] Here, \( a = \frac{\pi}{6} \) and \( b = \frac{\pi}{3} \). Their sum is \( a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi + 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \).
Applying the property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \), we get: \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan(\frac{\pi}{2} - x)}} \, dx \] Using the trigonometric identity \( \tan(\frac{\pi}{2} - x) = \cot x \): \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot x}} \, dx \] Rewrite \( \sqrt{\cot x} \) as \( \frac{1}{\sqrt{\tan x}} \): \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{1 + \frac{1}{\sqrt{\tan x}}} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{\frac{\sqrt{\tan x} + 1}{\sqrt{\tan x}}} \, dx \] \[ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx \quad \cdots (2) \] Now, we add equation (1) and equation (2): \[ I + I = \int_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} \, dx + \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx \] \[ 2I = \int_{\pi/6}^{\pi/3} \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} \, dx \] \[ 2I = \int_{\pi/6}^{\pi/3} 1 \, dx \] Evaluating the simple integral: \[ 2I = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6} \] Solving for I: \[ I = \frac{\pi}{12} \] Step 4: Final Answer:
We have successfully proven that \( I = \frac{\pi}{12} \).