Solving ∫ Arcsin(x) / (1+x²) Dx: A Step-by-Step Guide

Introduction

Hey guys! Today, we're diving deep into a fascinating integral problem: computing the definite integral of arcsin(x) / (1+x²) from 0 to 1. This integral, ∫[0 to 1] arcsin(x) / (1+x²) dx, might seem intimidating at first glance, but don't worry, we'll break it down step-by-step. We'll explore various techniques, from integration by parts to clever substitutions, and uncover the solution together. So, buckle up and let's embark on this mathematical journey!

When faced with such integrals, it's natural to feel a bit overwhelmed. The presence of both an inverse trigonometric function (arcsin(x)) and a rational function (1 / (1+x²)) hints at the need for strategic integration methods. We'll start by revisiting the initial attempt of integration by parts and understand why it didn't lead to a straightforward solution. Then, we'll explore alternative approaches, focusing on trigonometric substitutions and other techniques that might simplify the integral. The key is to identify the right substitution that transforms the integral into a more manageable form. This often involves recognizing patterns and relationships between the functions within the integral. By carefully choosing our substitutions, we can unravel the complexity and reveal the elegant solution hidden within. Remember, the beauty of calculus lies in its ability to transform seemingly difficult problems into solvable ones through creative techniques and a deep understanding of fundamental principles.

Initial Attempts: Integration by Parts

Let's kick things off by revisiting the initial approach: integration by parts. As a first instinct, it's a solid technique to try when you have a mix of function types. The formula for integration by parts is: ∫u dv = uv - ∫v du. The challenge lies in choosing the right 'u' and 'dv'. In our case, with ∫[0 to 1] arcsin(x) / (1+x²) dx, a natural choice for 'u' might seem to be arcsin(x), as its derivative simplifies things a bit. This makes 'dv' equal to dx / (1+x²). Let's see where this leads us.

So, if we let u = arcsin(x), then du = dx / √(1-x²). And if dv = dx / (1+x²), then v = arctan(x). Plugging these into the integration by parts formula, we get:

∫[0 to 1] arcsin(x) / (1+x²) dx = [arcsin(x) * arctan(x)] from 0 to 1 - ∫[0 to 1] arctan(x) / √(1-x²) dx.

Evaluating the first part, [arcsin(x) * arctan(x)] from 0 to 1, we get (π/2) * (π/4) - 0 = π²/8. So, our integral now looks like:

∫[0 to 1] arcsin(x) / (1+x²) dx = π²/8 - ∫[0 to 1] arctan(x) / √(1-x²) dx.

Now, here's where the initial attempt hits a snag. The new integral, ∫[0 to 1] arctan(x) / √(1-x²) dx, doesn't look any simpler than the original. In fact, it appears quite challenging on its own. This is a crucial moment in problem-solving – recognizing when a particular approach isn't bearing fruit. It's not a failure, but rather valuable information that guides us to explore other avenues. The complexity of this new integral suggests that integration by parts, while a reasonable first step, might not be the most efficient path to the solution. It's time to think outside the box and consider alternative techniques that might better suit the structure of the integral.

Exploring Trigonometric Substitution

Since integration by parts didn't immediately solve our problem, let's shift our focus to another powerful technique: trigonometric substitution. This method often shines when dealing with expressions involving square roots of the form √(a² - x²) or 1 + x² in the denominator, which is precisely what we have in our integral, ∫[0 to 1] arcsin(x) / (1+x²) dx. The key idea behind trigonometric substitution is to replace 'x' with a trigonometric function that simplifies the expression and, hopefully, the entire integral.

In our case, the presence of 1 + x² in the denominator suggests the substitution x = tan(θ). This is a classic trigonometric substitution choice, as it leverages the trigonometric identity 1 + tan²(θ) = sec²(θ), which will help simplify the denominator. Let's see how this substitution transforms our integral.

If we let x = tan(θ), then dx = sec²(θ) dθ. Also, arcsin(x) becomes arcsin(tan(θ)). We need to change the limits of integration as well. When x = 0, tan(θ) = 0, so θ = 0. When x = 1, tan(θ) = 1, so θ = π/4. Now we can rewrite the integral in terms of θ:

∫[0 to 1] arcsin(x) / (1+x²) dx = ∫[0 to π/4] arcsin(tan(θ)) / (1 + tan²(θ)) * sec²(θ) dθ.

Using the identity 1 + tan²(θ) = sec²(θ), the sec²(θ) terms cancel out, leaving us with:

∫[0 to π/4] arcsin(tan(θ)) dθ.

This looks much simpler! We've successfully transformed the integral into one involving a single function: arcsin(tan(θ)). However, we're not quite there yet. The arcsin(tan(θ)) term still needs some attention. The next step is to find a way to handle this composition of functions. We might need another clever substitution or perhaps a trigonometric identity to further simplify the integrand. The important thing is that we've made significant progress by employing trigonometric substitution, and we're now in a much better position to tackle the remaining challenge.

A Crucial Substitution: Letting θ = π/4 - u

Okay, guys, we've arrived at a pivotal moment in solving this integral. We've successfully transformed it into ∫[0 to π/4] arcsin(tan(θ)) dθ using the trigonometric substitution x = tan(θ). However, the arcsin(tan(θ)) term is still a bit tricky. This is where a touch of mathematical ingenuity comes into play. We're going to introduce another substitution, a subtle but powerful one: let θ = π/4 - u. This might seem like it's coming out of left field, but trust me, it's a game-changer!

Why this substitution? Well, the interval of integration [0, π/4] and the presence of tan(θ) hint that this substitution might exploit some trigonometric symmetries. By replacing θ with π/4 - u, we're essentially reflecting the interval around its midpoint. This often leads to simplifications, especially when dealing with trigonometric functions.

Let's see how this substitution works. If θ = π/4 - u, then dθ = -du. We also need to change the limits of integration. When θ = 0, u = π/4. When θ = π/4, u = 0. So, our integral becomes:

∫[0 to π/4] arcsin(tan(θ)) dθ = -∫[π/4 to 0] arcsin(tan(π/4 - u)) du.

We can flip the limits of integration and get rid of the negative sign:

∫[0 to π/4] arcsin(tan(θ)) dθ = ∫[0 to π/4] arcsin(tan(π/4 - u)) du.

Now, we need to deal with the tan(π/4 - u) term. Remember the tangent subtraction formula? tan(a - b) = (tan(a) - tan(b)) / (1 + tan(a)tan(b)). Applying this, we get:

tan(π/4 - u) = (tan(π/4) - tan(u)) / (1 + tan(π/4)tan(u)) = (1 - tan(u)) / (1 + tan(u)).

So, our integral now looks like:

∫[0 to π/4] arcsin((1 - tan(u)) / (1 + tan(u))) du.

This might still look a bit complicated, but we've made significant progress. The key is that we've transformed the argument of the arcsin function into a more manageable form. This new form hints at a potential connection to another trigonometric identity or a further simplification. The beauty of this substitution lies in its ability to reveal hidden structures within the integral, paving the way for the final steps of the solution. Hang in there, guys, we're getting closer!

Unveiling the Solution: Exploiting Symmetry

Alright, let's keep the momentum going! We've arrived at the integral ∫[0 to π/4] arcsin((1 - tan(u)) / (1 + tan(u))) du after our clever substitution θ = π/4 - u. Now, the next step is to recognize a crucial symmetry that will lead us to the solution. This is where mathematical intuition and pattern recognition become invaluable.

Let's denote our original integral as I: I = ∫[0 to π/4] arcsin(tan(θ)) dθ. We also know that I = ∫[0 to π/4] arcsin((1 - tan(u)) / (1 + tan(u))) du. Since 'u' is just a variable of integration, we can replace it with 'θ' without changing the value of the integral. So, we have:

I = ∫[0 to π/4] arcsin((1 - tan(θ)) / (1 + tan(θ))) dθ.

Now, let's add the two expressions for I together:

2I = ∫[0 to π/4] arcsin(tan(θ)) dθ + ∫[0 to π/4] arcsin((1 - tan(θ)) / (1 + tan(θ))) dθ

2I = ∫[0 to π/4] [arcsin(tan(θ)) + arcsin((1 - tan(θ)) / (1 + tan(θ)))] dθ.

This is where the magic happens! We need to simplify the expression inside the integral: arcsin(tan(θ)) + arcsin((1 - tan(θ)) / (1 + tan(θ))). Let's call this expression A. So,

A = arcsin(tan(θ)) + arcsin((1 - tan(θ)) / (1 + tan(θ))).

Recall that (1 - tan(θ)) / (1 + tan(θ)) = tan(π/4 - θ). So, we have:

A = arcsin(tan(θ)) + arcsin(tan(π/4 - θ)).

Now, let α = arcsin(tan(θ)) and β = arcsin(tan(π/4 - θ)). Then, tan(α) = tan(θ) and tan(β) = tan(π/4 - θ). We want to find α + β.

Here's another clever trick: consider the tangent addition formula again: tan(α + β) = (tan(α) + tan(β)) / (1 - tan(α)tan(β)).

Substituting our values, we get:

tan(α + β) = (tan(θ) + tan(π/4 - θ)) / (1 - tan(θ)tan(π/4 - θ)).

Now, substitute tan(π/4 - θ) = (1 - tan(θ)) / (1 + tan(θ)):

tan(α + β) = [tan(θ) + (1 - tan(θ)) / (1 + tan(θ))] / [1 - tan(θ)(1 - tan(θ)) / (1 + tan(θ))].

Simplifying this expression (which involves a bit of algebraic manipulation), we find that tan(α + β) = 1.

Since α and β are angles in the range [-π/2, π/2], α + β must be π/4. Therefore, A = α + β = π/4.

So, our integral simplifies dramatically:

2I = ∫[0 to π/4] (π/4) dθ.

2I = (π/4) * θ from 0 to π/4

2I = (π/4) * (π/4) = π²/16.

Finally, we can solve for I:

I = π²/32.

And there you have it, guys! The integral ∫[0 to 1] arcsin(x) / (1+x²) dx equals π²/32. We solved it by combining trigonometric substitution, a clever substitution (θ = π/4 - u), and a beautiful exploitation of symmetry. What a journey!

Conclusion

Wow, what a ride! We've successfully navigated the complexities of the integral ∫[0 to 1] arcsin(x) / (1+x²) dx and arrived at the elegant solution: π²/32. This problem beautifully illustrates the power and versatility of calculus techniques. We saw how integration by parts, while a natural first attempt, didn't lead to a direct solution. This prompted us to explore trigonometric substitution, which proved to be a key step in simplifying the integral. But the real magic happened with the substitution θ = π/4 - u, which unveiled a hidden symmetry that allowed us to reduce the integral to a simple form.

This journey highlights the importance of persistence and adaptability in problem-solving. It's not always about finding the right technique immediately, but rather about exploring different avenues, learning from our attempts, and creatively combining methods. The substitution θ = π/4 - u, in particular, demonstrates the power of thinking outside the box and looking for subtle relationships within the problem. The final step, where we exploited the symmetry by adding two different expressions for the integral, is a classic example of mathematical elegance and ingenuity.

So, the next time you encounter a challenging integral, remember the lessons we've learned here. Don't be afraid to experiment, explore different techniques, and look for hidden symmetries. With patience, creativity, and a solid understanding of calculus principles, you too can conquer even the most daunting integrals. Keep practicing, keep exploring, and keep the mathematical spirit alive! You've got this!