Key Concepts and Formulas

• Property of Definite Integrals: For a continuous function $f(x)$ on $[a, b]$, $\int\limits_a^b f(x) \, dx = \int\limits_a^b f(a+b-x) \, dx$. This property is particularly useful when the integrand transforms into a simpler form upon the substitution $x \to a+b-x$.
• Trigonometric Identities: $\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\sin(\frac{\pi}{2} - \theta) = \cos \theta$, $\cos(\frac{\pi}{2} - \theta) = \sin \theta$.

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Step-by-Step Solution

Step 1: Rewrite the Integrand

Let the given integral be $I$. $I = \int\limits_{\pi /24}^{5\pi /24} {{{dx} \over {1 + \root 3 \of {\tan 2x} }}}$ We rewrite the integrand in terms of sine and cosine. $\tan 2x = \frac{\sin 2x}{\cos 2x}$. So, $\root 3 \of {\tan 2x} = \left(\frac{\sin 2x}{\cos 2x}\right)^{1/3} = \frac{(\sin 2x)^{1/3}}{(\cos 2x)^{1/3}}$. Substituting this into the integral: $I = \int\limits_{\pi /24}^{5\pi /24} {{{dx} \over {1 + \frac{(\sin 2x)^{1/3}}{(\cos 2x)^{1/3}}}}}$ To simplify the denominator, we find a common denominator: $1 + \frac{(\sin 2x)^{1/3}}{(\cos 2x)^{1/3}} = \frac{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}}{(\cos 2x)^{1/3}}$ Now, the integral becomes: $I = \int\limits_{\pi /24}^{5\pi /24} {{{dx} \over {\frac{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}}{(\cos 2x)^{1/3}}}}}$ This simplifies to: $I = \int\limits_{\pi /24}^{5\pi /24} {{{(\cos 2x)^{1/3}} \over {{(\cos 2x)^{1/3}} + {{(\sin 2x)^{1/3}}}}}dx} \quad \text{...(1)}$ This form of the integrand is suitable for applying the property of definite integrals.

Step 2: Apply the Property of Definite Integrals

We use the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$. Here, $a = \frac{\pi}{24}$ and $b = \frac{5\pi}{24}$. The sum of the limits is $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$. We replace $x$ with $(a+b-x)$, which is $(\frac{\pi}{4} - x)$. The argument of the trigonometric functions is $2x$. So, we replace $2x$ with $2(\frac{\pi}{4} - x) = \frac{\pi}{2} - 2x$. Let's see how the terms in the integrand transform: $\cos(2x) \to \cos(\frac{\pi}{2} - 2x) = \sin(2x)$ $\sin(2x) \to \sin(\frac{\pi}{2} - 2x) = \cos(2x)$ Applying this transformation to Equation (1): $I = \int\limits_{\pi /24}^{5\pi /24} {{{(\sin 2x)^{1/3}} \over {{(\sin 2x)^{1/3}} + {{(\cos 2x)^{1/3}}}}}dx} \quad \text{...(2)}$ This new integral, Equation (2), represents the same value as the original integral $I$.

Step 3: Add the Two Integrals

Now, we add Equation (1) and Equation (2): $I + I = \int\limits_{\pi /24}^{5\pi /24} {{{(\cos 2x)^{1/3}} \over {{(\cos 2x)^{1/3}} + {{(\sin 2x)^{1/3}}}}}dx} + \int\limits_{\pi /24}^{5\pi /24} {{{(\sin 2x)^{1/3}} \over {{(\sin 2x)^{1/3}} + {{(\cos 2x)^{1/3}}}}}dx}$ Since the limits of integration are the same, we can combine the integrands: $2I = \int\limits_{\pi /24}^{5\pi /24} {\left( {{{(\cos 2x)^{1/3}} \over {{(\cos 2x)^{1/3}} + {{(\sin 2x)^{1/3}}}}} + {{{(\sin 2x)^{1/3}}} \over {{(\sin 2x)^{1/3}} + {{(\cos 2x)^{1/3}}}}}} \right)} dx$ The denominators are identical, so we add the numerators: $2I = \int\limits_{\pi /24}^{5\pi /24} {\left( {{{(\cos 2x)^{1/3}} + {{(\sin 2x)^{1/3}}} \over {{(\cos 2x)^{1/3}} + {{(\sin 2x)^{1/3}}}}}} \right)} dx$ The numerator and the denominator are the same, so the integrand simplifies to $1$: $2I = \int\limits_{\pi /24}^{5\pi /24} 1 \, dx$

Step 4: Evaluate the Simplified Integral

Now we evaluate the integral of $1$ with respect to $x$: $2I = [x]_{\pi /24}^{5\pi /24}$ $2I = \frac{5\pi}{24} - \frac{\pi}{24}$ $2I = \frac{4\pi}{24}$ $2I = \frac{\pi}{6}$ Solving for $I$: $I = \frac{1}{2} \times \frac{\pi}{6}$ $I = \frac{\pi}{12}$

Common Mistakes & Tips

• Incorrect Substitution: Ensure that when applying the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, you correctly substitute for the entire argument of the trigonometric functions, not just $x$. In this case, $2x$ was replaced by $2(a+b-x)$.
• Algebraic Errors: Be meticulous when rewriting the integrand and combining fractions to avoid algebraic mistakes.
• Forgetting to Divide by 2: After finding $2I$, remember to divide by 2 to get the value of $I$.

Summary

The problem involves a definite integral where the integrand is of the form $f(x) = \frac{1}{1 + \sqrt[3]{\tan 2x}}$. By rewriting the integrand in terms of sine and cosine, we obtained $f(x) = \frac{(\cos 2x)^{1/3}}{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}}$. Applying the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ with $a=\pi/24$ and $b=5\pi/24$ (so $a+b=\pi/4$), the integrand transformed to $\frac{(\sin 2x)^{1/3}}{(\sin 2x)^{1/3} + (\cos 2x)^{1/3}}$. Adding the original and the transformed integral resulted in an integrand of $1$. Evaluating $\int_{\pi/24}^{5\pi/24} 1 \, dx$ gave $2I = \pi/6$, leading to $I = \pi/12$.

The final answer is $\boxed{{\pi \over {12}}}$.