Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{\pi /24}^{5\pi /24}\frac{dx}{1+\sqrt[3]{\tan 2x}}$ is :

Mastering the King's Property

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving an integral; we are uncovering a hidden symmetry.

When you first look at the integral

it is natural to feel a sense of intimidation. The cube root of a tangent function looks like a trap designed to waste your time, but in the world of JEE Advanced, complexity is often just a mask for elegance.

Recognizing the Pattern

Before we touch a single piece of chalk, we must observe the boundaries. Our limits are $a=\frac{\pi }{24}$ and $b=\frac{5\pi }{24}$.

Notice anything special? Their sum is $a+b=\frac{6\pi }{24}=\frac{\pi }{4}$.

Whenever you see limits that sum to a value like $\frac{\pi }{4}$ or $\frac{\pi }{2}$ in a trigonometric integral, your internal alarm should ring. This is the signature of the King's Property—the most powerful tool in your integration arsenal.

The King's Property states that for any continuous function $f(x)$, the integral ${\int }_a^{b}f(x)dx$ is identical to ${\int }_a^{b}f(a+b-x)dx$. We are essentially 'flipping' the function across the midpoint of the interval.

Let us apply this to our integral by substituting $x$ with $\frac{\pi }{4}-x$. This transforms our argument $2x$ into $2(\frac{\pi }{4}-x)=\frac{\pi }{2}-2x$.

Now, recall the beauty of trigonometry: $\tan (\frac{\pi }{2}-2x)=\cot 2x$. Suddenly, our integral transforms into:

We are now standing at the threshold of the solution. We have two versions of the same integral $I$.

If we express $\cot 2x$ as $\frac{1}{\tan 2x}$, we get:

By multiplying the numerator and denominator by $\sqrt[3]{\tan 2x}$, we arrive at:

Now, watch the magic happen. When we add the original $I$ to this new version, the denominators are identical! We get:

The term inside the integral becomes $\frac{1+\sqrt[3]{\tan 2x}}{1+\sqrt[3]{\tan 2x}}$, which is simply $1$. The entire complex expression has vanished, leaving us with the humble integral of a constant.

We are left with $2I={\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}1dx$.

This is a trivial calculation:

Dividing by $2$, we find the final answer:

Look at what we have achieved. We didn't fight the function; we danced with it. We used its own symmetry to simplify it until it had no choice but to reveal its answer. This is the essence of JEE Advanced mathematics—not brute force, but the refined application of fundamental principles.