Question

The value of the integral \( \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1 + \sqrt[3]{\tan 2x}} \) is :

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{18} \)
• C. \( \frac{\pi}{12} \)
• D. \( \frac{\pi}{3} \)

Hint:

For integrals of the form \( \frac{1}{1+\tan^n x} \) with limits \( a, b \), the answer is \( \frac{b-a}{2} \) if \( a+b = \pi/2 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem is solved using the property \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \). In this case, the sum of the limits \( \pi/24 + 5\pi/24 = \pi/4 \).
Step 2: Key Formula or Approach:
Apply King's Property: \( I = \int_a^b \frac{1}{1 + (f(x))^n} dx \).
Step 3: Detailed Explanation:
Let \( I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\tan 2x}} \). Sum of limits \( = \pi/4 \). Replace \( x \) with \( \pi/4 - x \), so \( 2x \) becomes \( \pi/2 - 2x \). Since \( \tan(\pi/2 - 2x) = \cot 2x \): \[ I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\cot 2x}} = \int_{\pi/24}^{5\pi/24} \frac{\sqrt[3]{\tan 2x}}{1 + \sqrt[3]{\tan 2x}} dx \] Adding both equations: \[ 2I = \int_{\pi/24}^{5\pi/24} 1 \cdot dx = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6} \] \[ I = \frac{\pi}{12} \] Step 4: Final Answer:
The integral value is \( \frac{\pi}{12} \).