Solving Two Tricky Ahmed Integral Variations

Introduction

Hey guys! Today, we're diving into the fascinating world of calculus, specifically exploring two intriguing variations of the Ahmed integral. These integrals, involving trigonometric and algebraic functions, might seem daunting at first, but don't worry, we'll break them down step by step. We'll be focusing on the integrals:

$$\int_{\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{2}}} \frac{\arctan(\sqrt{5}\sqrt{1-2x^2})}{1+x^2} \text{d}x$$

and another related integral. Our main goal is to find closed-form solutions for these integrals, meaning we want to express the results in terms of elementary functions and constants. This journey will involve clever substitutions, trigonometric identities, and a bit of mathematical ingenuity. So, buckle up and let's get started!

Diving Deep into Ahmed Integrals

Ahmed integrals, in general, are a class of definite integrals that often involve inverse trigonometric functions and algebraic expressions. These integrals are known for their non-trivial solutions and often require a combination of techniques to solve. The specific integrals we're tackling today are particularly interesting because they showcase how seemingly small changes in the integrand or limits of integration can lead to significant differences in the solution process.

Before we jump into the nitty-gritty, let's talk about why these integrals are so captivating. First off, they beautifully blend trigonometric functions like arctan with algebraic functions, creating a delightful challenge for anyone who loves calculus. When we encounter $\arctan(\sqrt{5}\sqrt{1-2x^2})$, it's like spotting a unique puzzle piece – we know we'll need to use some cool tricks to unravel it. Plus, these integrals often pop up in various fields of mathematics and physics, making them not just academic exercises but valuable tools in real-world applications.

Our mission here is to nail down those closed-form solutions. This means we're aiming for answers that are neat and tidy, expressed using familiar functions and constants. Think of it as finding the perfect, polished answer rather than a messy approximation. To get there, we’ll be rolling up our sleeves and diving into substitutions, those clever moves where we swap one variable for another to simplify the integral. Trigonometric identities will be our trusty sidekicks, helping us reshape expressions into more manageable forms. And, of course, a dash of mathematical creativity will be essential – because sometimes, the best solutions come from thinking outside the box.

The First Integral: A Detailed Exploration

Let's begin with our first integral:

$$\int_{\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{2}}} \frac{\arctan(\sqrt{5}\sqrt{1-2x^2})}{1+x^2} \text{d}x$$

This integral looks complex, but let's break it down. The key here is to recognize the structure of the integrand. We have an arctangent function composed with a square root term, all divided by $1 + x^2$. This structure hints at a trigonometric substitution. A common strategy in such cases is to substitute $x = \tan(\theta)$, which simplifies the $1 + x^2$ term in the denominator.

Trigonometric Substitution: When we see a mix of algebraic and trigonometric functions, especially with square roots, trigonometric substitutions are our go-to move. For this integral, substituting $x = \tan(\theta)$ is like fitting the right key into the lock. It transforms the often-awkward $1 + x^2$ into a much friendlier trigonometric expression. This substitution is a total game-changer because it simplifies the integral's structure, making it easier to handle. Think of it as turning a tangled knot into a neat, orderly loop.

Transforming the Limits: As we switch variables, we need to adjust the limits of integration to match our new variable, $\theta$. This step is crucial because it ensures our solution stays accurate. We're not just changing the expression inside the integral; we're also redefining the boundaries over which we're integrating. For the lower limit, $x = \frac{1}{\sqrt{3}}$, we find $\theta = \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$. For the upper limit, $x = \frac{1}{\sqrt{2}}$, we get $\theta = \arctan(\frac{1}{\sqrt{2}})$. These new limits are our guideposts, ensuring we're integrating over the correct range in terms of $\theta$.

Simplifying the Integrand: Now, the fun part – simplifying the integral. When we substitute $x = \tan(\theta)$, we also have $dx = \sec^2(\theta) d\theta$. This little transformation is super important because it links our original variable, $x$, to our new variable, $\theta$. After plugging in these substitutions, the integral starts to look more manageable. The $1 + x^2$ in the denominator becomes $1 + \tan^2(\theta)$, which conveniently simplifies to $\sec^2(\theta)$. The square root term, $\sqrt{1 - 2x^2}$, transforms into $\sqrt{1 - 2\tan^2(\theta)}$. This is where things get interesting – we're reshaping the integral piece by piece, turning a complex puzzle into something we can actually solve.

After the substitution, the integral becomes:

$$\int_{\frac{\pi}{6}}^{\arctan(\frac{1}{\sqrt{2}})} \frac{\arctan(\sqrt{5}\sqrt{1-2\tan^2(\theta)})}{1+\tan^2(\theta)} \sec^2(\theta) \text{d}\theta$$

Using the identity $1 + \tan^2(\theta) = \sec^2(\theta)$, we can simplify further:

$$\int_{\frac{\pi}{6}}^{\arctan(\frac{1}{\sqrt{2}})} \arctan(\sqrt{5}\sqrt{1-2\tan^2(\theta)}) \text{d}\theta$$

This simplified form is a significant step forward, but we're not quite there yet. The term inside the arctangent still needs some love. We'll need to employ further trigonometric identities and potentially another substitution to fully crack this integral.

The Other Integral: A Comparative Analysis

Now, let's consider the