Solving The Integral Of Arccos(x)f(x^2)/x A Step-by-Step Guide

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Hey guys! Today, we're diving headfirst into a fascinating integral problem that looks intimidating at first glance, but trust me, it’s a super rewarding journey once we break it down. Our mission, should we choose to accept it, is to prove that the definite integral $\int^1_0 \frac{\arccos(x)}{x}f(x^2) dx$ equals $\frac{\pi^4}{512}$, where $f(x)$ is a pretty wild-looking function: $f(x)=\arctan\left({\sqrt{\frac{\sqrt{1+x^2}-1}{x}}}\right)\text{arctanh}\left({\sqrt{\frac{\sqrt{1+x^2}-1}{x}}}\right)$. Buckle up, because we're about to embark on a mathematical adventure!

1. Unpacking the Challenge: Understanding the Integral

When we first encounter an integral like this, it’s natural to feel a bit overwhelmed. Let's break down the key components to make it more digestible. The integral $\int^1_0 \frac{\arccos(x)}{x}f(x^2) dx$ is a definite integral, which means we're calculating the area under the curve of the function $\frac{\arccos(x)}{x}f(x^2)$ between the limits of integration, 0 and 1.

The heart of the integral lies in the function $f(x^2)$. Remember, $f(x)$ is defined as $\arctan\left({\sqrt{\frac{\sqrt{1+x^2}-1}{x}}}\right)\text{arctanh}\left({\sqrt{\frac{\sqrt{1+x^2}-1}{x}}}\right)$. So, $f(x^2)$ means we're plugging in $x^2$ wherever we see $x$ in the expression for $f(x)$. This gives us:

$f(x^2) = \arctan\left({\sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}}\right)\text{arctanh}\left({\sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}}\right)$

The presence of both $\arctan$ (arctangent) and $\text{arctanh}$ (inverse hyperbolic tangent) functions suggests we might need to employ some trigonometric and hyperbolic identities to simplify things. The nested square roots further hint at the possibility of using clever substitutions to unravel the expression. The $\arccos(x)$ term adds another layer of trigonometric complexity. It represents the inverse cosine function, which gives us the angle whose cosine is $x$. This means we will need to recall our knowledge of inverse trigonometric functions and their properties. To solve the integral, we'll need a strategy. Common techniques for tackling definite integrals include: u-substitution, integration by parts, trigonometric substitution, and sometimes, even contour integration (though that's more advanced). Given the structure of our integral, especially the composite functions within $f(x^2)$, a well-chosen substitution seems like a promising starting point. We might also need to explore trigonometric and hyperbolic identities to simplify the expression inside the integral. The goal here is to transform the integral into a form that we can evaluate using standard techniques or known results.

2. The Substitution Game: Simplifying the Beast

The key to cracking this integral open lies in a strategic substitution. Let's consider the inner part of our f(x) function, specifically the $\sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}$ term within f(x²). This looks like a prime candidate for simplification. Let's try the substitution:

$u = \sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}$

Squaring both sides, we get:

$u^2 = \frac{\sqrt{1+x^4}-1}{x^2}$

Now, let’s isolate the square root term:

$u^2x^2 = \sqrt{1+x^4} - 1$

$u^2x^2 + 1 = \sqrt{1+x^4}$

Squaring again gives us:

$(u^2x^2 + 1)^2 = 1 + x^4$

Expanding the left side:

$u^4x^4 + 2u^2x^2 + 1 = 1 + x^4$

This simplifies to:

$u^4x^4 + 2u^2x^2 = x^4$

Now, let's factor out $x^2$:

$x^2(u^4x^2 + 2u^2) = x^4$

If $x \neq 0$ (which is fine in our interval of integration), we can divide both sides by $x^2$:

$u^4x^2 + 2u^2 = x^2$

Rearranging to solve for $x^2$:

$x^2(u^4 - 1) = -2u^2$

$x^2 = \frac{2u^2}{1 - u^4}$

This gives us an expression for $x^2$ in terms of u. Now, let's find an expression for x:

$x = \sqrt{\frac{2u^2}{1 - u^4}} = u\sqrt{\frac{2}{1 - u^4}}$

Next, we need to find dx in terms of du. This is where things get a little hairy, but we can handle it. Differentiating $x^2 = \frac{2u^2}{1 - u^4}$ with respect to u, we get:

$2x \frac{dx}{du} = \frac{(1 - u^4)(4u) - 2u^2(-4u^3)}{(1 - u^4)^2}$

$2x \frac{dx}{du} = \frac{4u - 4u^5 + 8u^5}{(1 - u^4)^2}$

$2x \frac{dx}{du} = \frac{4u + 4u^5}{(1 - u^4)^2}$

$\frac{dx}{du} = \frac{2u(1 + u^4)}{x(1 - u^4)^2}$

Substituting the expression for x we found earlier:

$\frac{dx}{du} = \frac{2u(1 + u^4)}{u\sqrt{\frac{2}{1 - u^4}}(1 - u^4)^2}$

$\frac{dx}{du} = \frac{2(1 + u^4)}{\sqrt{2}(1 - u^4)^{3/2}}$

So, $dx = \frac{2(1 + u^4)}{\sqrt{2}(1 - u^4)^{3/2}} du$

Now, let's consider the limits of integration. When $x = 0$, we have $u = \sqrt{\frac{\sqrt{1+0}-1}{0}}$, which is an indeterminate form. We'll need to take a limit as $x$ approaches 0. When x = 1, u = $\sqrt{\frac{\sqrt{1+1}-1}{1}} = \sqrt{\sqrt{2}-1}$. The substitution seems complex, but it transforms f(x²) into a much simpler form: $f(x^2) = \arctan(u)\text{arctanh}(u)$. This is a significant simplification! However, the dx term and the change of limits are quite involved, indicating that perhaps a different approach or a further simplification of this substitution might be necessary. Let's hold onto this for now and see if other avenues open up.

3. A Trigonometric Turn: Exploring Arccosine and Trigonometric Identities

Let's shift our focus to the $\arccos(x)$ term in the integral. Since it's an inverse trigonometric function, it's worth exploring if a trigonometric substitution could simplify things. A natural substitution here is:

$x = \cos(\theta)$

This implies that $\arccos(x) = \theta$. Now we need to find dx in terms of dθ:

$dx = -\sin(\theta) d\theta$

We also need to change the limits of integration. When $x = 0$, $\cos(\theta) = 0$, which means $\theta = \frac{\pi}{2}$. When $x = 1$, $\cos(\theta) = 1$, so $\theta = 0$. Therefore, our integral transforms to:

$\int^1_0 \frac{\arccos(x)}{x}f(x^2) dx = \int_{\pi/2}^0 \frac{\theta}{\cos(\theta)} f(\cos^2(\theta))(-\sin(\theta)) d\theta$

We can reverse the limits of integration and change the sign:

$= \int_0^{\pi/2} \frac{\theta}{\cos(\theta)} f(\cos^2(\theta)) \sin(\theta) d\theta$

$= \int_0^{\pi/2} \theta \tan(\theta) f(\cos^2(\theta)) d\theta$

Now, we need to express $f(\cos^2(\theta))$ in terms of $\theta$. Recall that $f(x) = \arctan\left({\sqrt{\frac{\sqrt{1+x^2}-1}{x}}}\right)\text{arctanh}\left({\sqrt{\frac{\sqrt{1+x^2}-1}{x}}}\right)$. So,

$f(\cos^2(\theta)) = \arctan\left({\sqrt{\frac{\sqrt{1+\cos^4(\theta)}-1}{\cos^2(\theta)}}}\right)\text{arctanh}\left({\sqrt{\frac{\sqrt{1+\cos^4(\theta)}-1}{\cos^2(\theta)}}}\right)$

This looks complicated, but let’s see if we can simplify the expression inside the square roots. This trigonometric substitution has simplified the arccos(x) part nicely, but it has made the f term even more complex. However, we now have a tan(θ) term, which might be useful. Let's take a closer look at the expression inside the arctan and arctanh functions within f(cos²(θ)) and see if we can use any trigonometric identities. This might involve half-angle formulas or other trigonometric manipulations. The goal is to simplify the nested radicals and hopefully reveal a more manageable form.

4. Half-Angle Harmony: Taming the Trigonometric Term

Let’s zoom in on the expression inside the square root within f(cos²(θ)): $\sqrt{\frac{\sqrt{1+\cos^4(\theta)}-1}{\cos^2(\theta)}}$. To simplify this, we can try to manipulate the $\sqrt{1+\cos^4(\theta)}$ term. However, a more promising approach might be to use a half-angle identity. Let's try to relate this expression to a tangent half-angle formula. Recall the identity:

$\tan^2(\frac{x}{2}) = \frac{1 - \cos(x)}{1 + \cos(x)}$

We want to make the expression inside our square root look similar to this. Notice that if we let $x = 2\alpha$, then $\tan^2(\alpha) = \frac{1 - \cos(2\alpha)}{1 + \cos(2\alpha)}$. This suggests we should look for a way to express our term in a similar form. Let's go back to our substitution $x = \cos(\theta)$ and the expression $u = \sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}$. We had $x^2 = \frac{2u^2}{1 - u^4}$. Now, let's try a different trigonometric approach directly on u. Let's set $u = \tan(\alpha)$. Then:

$\tan^2(\alpha) = \frac{\sqrt{1+x^4}-1}{x^2}$

$\tan^2(\alpha)x^2 = \sqrt{1+x^4} - 1$

$\tan^2(\alpha)x^2 + 1 = \sqrt{1+x^4}$

Squaring both sides:

$(\tan^2(\alpha)x^2 + 1)^2 = 1 + x^4$

$\tan^4(\alpha)x^4 + 2\tan^2(\alpha)x^2 + 1 = 1 + x^4$

$\tan^4(\alpha)x^4 + 2\tan^2(\alpha)x^2 = x^4$

If $x \neq 0$, divide by $x^2$:

$\tan^4(\alpha)x^2 + 2\tan^2(\alpha) = x^2$

$x^2(1 - \tan^4(\alpha)) = 2\tan^2(\alpha)$

$x^2 = \frac{2\tan^2(\alpha)}{1 - \tan^4(\alpha)}$

Now, recall the double angle formula for tangent: $\tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)}$. We can rewrite our expression for $x^2$ as:

$x^2 = \frac{2\tan^2(\alpha)}{(1 - \tan^2(\alpha))(1 + \tan^2(\alpha))}$

Multiply the numerator and denominator by $(1 - \tan^2(\alpha))$:

$x^2 = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \cdot \frac{\tan(\alpha)}{1 + \tan^2(\alpha)}$

$x^2 = \tan(2\alpha) \frac{\tan(\alpha)}{1 + \tan^2(\alpha)}$

This is progress! Now, let’s think back to our original function f(x). With the substitution $u = \tan(\alpha)$, we have:

$f(x^2) = \arctan(u)\text{arctanh}(u) = \arctan(\tan(\alpha))\text{arctanh}(\tan(\alpha)) = \alpha \text{arctanh}(\tan(\alpha))$

This is a crucial simplification! We've managed to express f(x²) in terms of α. However, we still need to relate α back to θ (from our x = cos(θ) substitution) and express the entire integral in terms of α. This might involve using the relationship we derived between x² and tan(α), along with some trigonometric manipulation. The appearance of $\text{arctanh}(\tan(\alpha))$ is interesting, suggesting a possible connection between hyperbolic and trigonometric functions that we can exploit. We need to keep pushing forward, connecting the pieces we've found to create a complete solution.

5. The Final Flourish: Putting It All Together

This is where we bring all our hard work together. We have several key relationships:

• $x = \cos(\theta)$ (from our trigonometric substitution)
• $u = \sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}$ (our initial substitution within f)
• $u = \tan(\alpha)$ (our second trigonometric substitution)
• $f(x^2) = \alpha \text{arctanh}(\tan(\alpha))$
• $x^2 = \frac{2\tan^2(\alpha)}{1 - \tan^4(\alpha)}$

Our integral, after the x = cos(θ) substitution, became:

$\int_0^{\pi/2} \theta \tan(\theta) f(\cos^2(\theta)) d\theta$

We need to express everything in terms of α. We know $f(\cos^2(\theta)) = \alpha \text{arctanh}(\tan(\alpha))$. The challenge now is to express θ and tan(θ) in terms of α. From $x = \cos(\theta)$ and $x^2 = \frac{2\tan^2(\alpha)}{1 - \tan^4(\alpha)}$, we have:

$\cos^2(\theta) = \frac{2\tan^2(\alpha)}{1 - \tan^4(\alpha)}$

This gives us a relationship between θ and α, but it's still quite complex. Let's try a different approach. We had the substitution $u = \tan(\alpha) = \sqrt{\frac{\sqrt{1+x^4}-1}{x^2}}$. Substituting $x = \cos(\theta)$ into this, we get:

$\tan(\alpha) = \sqrt{\frac{\sqrt{1+\cos^4(\theta)}-1}{\cos^2(\theta)}}$

This looks very similar to the expression we were trying to simplify earlier! Let's square both sides:

$\tan^2(\alpha) = \frac{\sqrt{1+\cos^4(\theta)}-1}{\cos^2(\theta)}$

$\tan^2(\alpha)\cos^2(\theta) = \sqrt{1+\cos^4(\theta)}-1$

$\tan^2(\alpha)\cos^2(\theta) + 1 = \sqrt{1+\cos^4(\theta)}$

Squaring again:

$(\tan^2(\alpha)\cos^2(\theta) + 1)^2 = 1 + \cos^4(\theta)$

$\tan^4(\alpha)\cos^4(\theta) + 2\tan^2(\alpha)\cos^2(\theta) + 1 = 1 + \cos^4(\theta)$

$\tan^4(\alpha)\cos^4(\theta) + 2\tan^2(\alpha)\cos^2(\theta) = \cos^4(\theta)$

If $\cos(\theta) \neq 0$, divide by $\cos^2(\theta)$:

$\tan^4(\alpha)\cos^2(\theta) + 2\tan^2(\alpha) = \cos^2(\theta)$

$\cos^2(\theta)(1 - \tan^4(\alpha)) = 2\tan^2(\alpha)$

$\cos^2(\theta) = \frac{2\tan^2(\alpha)}{1 - \tan^4(\alpha)}$

This is the same expression for $\cos^2(\theta)$ we derived earlier! Now, let's use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$:

$\sin^2(\theta) = 1 - \frac{2\tan^2(\alpha)}{1 - \tan^4(\alpha)}$

$\sin^2(\theta) = \frac{1 - \tan^4(\alpha) - 2\tan^2(\alpha)}{1 - \tan^4(\alpha)}$

Now we can find $\tan^2(\theta) = \frac{\sin^2(\theta)}{\cos^2(\theta)}$:

$\tan^2(\theta) = \frac{1 - \tan^4(\alpha) - 2\tan^2(\alpha)}{2\tan^2(\alpha)}$

This is still quite complicated. We've made significant progress in simplifying the function f(x²), but relating θ and α directly is proving to be challenging. The complexity of these relationships suggests that perhaps a more direct route exists, or that a special identity or property is being overlooked. Let's step back and review our progress, looking for alternative pathways or simplifications.

Given the difficulties in directly relating θ and α, it may be beneficial to explore integration by parts or consider special function properties related to arctan and arctanh. Another strategy might involve numerical integration techniques to approximate the value of the integral and verify if it indeed equals $\frac{\pi^4}{512}$. While this doesn't provide a formal proof, it can offer valuable insights and confirm the validity of our approach.

We have shown a comprehensive attempt to solve this integral, highlighting the common techniques and challenges encountered. While a complete closed-form solution hasn't been reached in this discussion, the exploration has revealed the intricate nature of the problem and the potential for further investigation. Keep exploring, keep questioning, and you'll eventually find the solution. Math is an exploration, not just a destination. You've got this!