Question

$\int_{\pi/6}^{\pi/3} \frac{\tan x}{\tan x + \cot x} dx$ is equal to

• A. $\frac{\pi}{4}$
• B. 0
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{12}$

Hint:

For definite integrals of the form \(\int_{a}^{b} \frac{f(x)}{f(x)+f(a+b-x)} dx\), the result is always \(\frac{b-a}{2}\). Here, \(f(x) = \tan x\), so \(f(a+b-x) = f(\frac{\pi}{2}-x) = \tan(\frac{\pi}{2}-x) = \cot x\). The integral is in this form, so the answer is \((\frac{\pi}{3} - \frac{\pi}{6})/2 = (\frac{\pi}{6})/2 = \frac{\pi}{12}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This definite integral can be solved efficiently using a property of definite integrals, often known as the "King's property".
Step 2: Key Formula or Approach:
The property states that \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\).
Here, \(a = \frac{\pi}{6}\) and \(b = \frac{\pi}{3}\). So, \(a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi + 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\).
Step 3: Detailed Explanation:
Let the given integral be I.
\[ I = \int_{\pi/6}^{\pi/3} \frac{\tan x}{\tan x + \cot x} dx \quad \cdots (1) \] Using the property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\), we replace x with \(a+b-x = \frac{\pi}{2}-x\).
\[ I = \int_{\pi/6}^{\pi/3} \frac{\tan(\frac{\pi}{2}-x)}{\tan(\frac{\pi}{2}-x) + \cot(\frac{\pi}{2}-x)} dx \] We know that \(\tan(\frac{\pi}{2}-x) = \cot x\) and \(\cot(\frac{\pi}{2}-x) = \tan x\).
\[ I = \int_{\pi/6}^{\pi/3} \frac{\cot x}{\cot x + \tan x} dx \quad \cdots (2) \] Now, add equation (1) and equation (2):
\[ I + I = \int_{\pi/6}^{\pi/3} \frac{\tan x}{\tan x + \cot x} dx + \int_{\pi/6}^{\pi/3} \frac{\cot x}{\cot x + \tan x} dx \] \[ 2I = \int_{\pi/6}^{\pi/3} \frac{\tan x + \cot x}{\tan x + \cot x} dx \] \[ 2I = \int_{\pi/6}^{\pi/3} 1 \cdot dx \] \[ 2I = [x]_{\pi/6}^{\pi/3} \] \[ 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6} \] \[ I = \frac{\pi}{12} \] Step 4: Final Answer:
The value of the integral is $\frac{\pi{12}$}.