Question

Prove that \(\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}} = \frac{\pi}{12}\).

Hint:

Whenever you see an integral of the form \(\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} dx\), its value is always \(\frac{b-a}{2}\). The given problem can be converted to this form by writing \(\tan x = \frac{\sin x}{\cos x}\), which makes this a very quick check. Here, \(a+b = \pi/2\), so \(f(a+b-x) = f(\pi/2 - x)\).

Answer 1

Step 1: Understanding the Concept:
This integral is a classic example that can be solved efficiently using a special property of definite integrals, often called the "King's property". This property is particularly useful when the integrand has a certain symmetry with respect to the sum of the limits.
Step 2: Key Formula or Approach:
We will use the property of definite integrals: \[ \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx \] The strategy is to apply this property to the given integral, add the new form of the integral to the original one, and simplify the result.
Step 3: Detailed Explanation:
Let the given integral be I: \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{\tan x}} \, dx \quad \text{---(1)} \] Here, \(a = \frac{\pi}{6}\) and \(b = \frac{\pi}{3}\). Their sum is \(a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi+2\pi}{6} = \frac{\pi}{2}\).
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 \(\frac{1}{\sqrt{\tan x}}\): \[ I = \int_{\pi/6}^{\pi/3} \frac{1}{1+\frac{1}{\sqrt{\tan x}}} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{\frac{\sqrt{\tan x}+1}{\sqrt{\tan x}}} \, dx \] \[ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} \, dx \quad \text{---(2)} \] Now, add equation (1) and equation (2): \[ I + I = \int_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{\tan x}} \, dx + \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} \, dx \] \[ 2I = \int_{\pi/6}^{\pi/3} \left( \frac{1 + \sqrt{\tan x}}{1+\sqrt{\tan x}} \right) \, dx \] \[ 2I = \int_{\pi/6}^{\pi/3} 1 \, dx \] Now, evaluate the simple integral: \[ 2I = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6} \] Finally, solve for I: \[ I = \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12} \] Step 4: Final Answer:
We have shown that the value of the integral is \(\frac{\pi}{12}\). Hence proved.