The value of the definite integral $\int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\tan 2x}}$ is :
Show Hint
For any integral of the form $\int_a^b \frac{dx}{1 + f(x)}$, if $f(a+b-x) = 1/f(x)$, the result is simply $\frac{b-a}{2}$. This specific pattern is very common with trigonometric functions where $a+b = \pi/2$.
Step 1: Understanding the Concept:
This definite integral can be simplified by first using a substitution and then applying the property \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\), often called the "King's Rule". Step 3: Detailed Explanation:
Let \(I = \int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\tan 2x}}\).
First, substitute \(2x = u \Rightarrow 2 dx = du \Rightarrow dx = du/2\).
Limits change: \(x = \pi/24 \to u = \pi/12\) and \(x = 5\pi/24 \to u = 5\pi/12\).
\[ I = \frac{1}{2} \int_{\pi/12}^{5\pi/12} \frac{du}{1 + \sqrt[3]{\tan u}} \]
Now apply King's property: \(a + b = \pi/12 + 5\pi/12 = 6\pi/12 = \pi/2\).
Replace \(u\) with \(\pi/2 - u\):
\[ I = \frac{1}{2} \int_{\pi/12}^{5\pi/12} \frac{du}{1 + \sqrt[3]{\tan(\pi/2 - u)}} = \frac{1}{2} \int_{\pi/12}^{5\pi/12} \frac{du}{1 + \sqrt[3]{\cot u}} \]
Convert \(\cot u\) to \(1/\tan u\):
\[ I = \frac{1}{2} \int_{\pi/12}^{5\pi/12} \frac{\sqrt[3]{\tan u}}{1 + \sqrt[3]{\tan u}} du \]
Add the two forms of \(I\):
\[ 2I = \frac{1}{2} \int_{\pi/12}^{5\pi/12} \left[ \frac{1}{1 + \sqrt[3]{\tan u}} + \frac{\sqrt[3]{\tan u}}{1 + \sqrt[3]{\tan u}} \right] du \]
\[ 2I = \frac{1}{2} \int_{\pi/12}^{5\pi/12} 1 \cdot du = \frac{1}{2} [u]_{\pi/12}^{5\pi/12} \]
\[ 2I = \frac{1}{2} \left( \frac{5\pi}{12} - \frac{\pi}{12} \right) = \frac{1}{2} \cdot \frac{4\pi}{12} = \frac{\pi}{6} \]
\[ I = \frac{\pi}{12} \] Step 4: Final Answer:
The value of the definite integral is \(\pi/12\).
