Integral (5-x^2)^(-3/2): Why X=√(5)sin(u)?

by Felix Dubois 43 views

Hey guys! Ever been stuck on an integral that just wouldn't budge? I totally get it. Integrals can be tricky beasts, especially when it comes to choosing the right substitution. Let's dive into a specific head-scratcher: the indefinite integral of (5-x2)(-3/2) over the interval [-1, 2]. Our mission? To figure out why the substitution x = √5sin(u) works like a charm, while other seemingly valid substitutions might leave us in the lurch. Buckle up, because we're about to embark on a journey through the land of trigonometric substitutions, domain restrictions, and the subtle art of choosing the right path.

The Core Challenge: Understanding the Integral

Let's start by dissecting the integral itself. We're dealing with ∫(5-x2)(-3/2) dx. Notice the form: a constant squared (5) minus a variable squared (x^2), all raised to a power. This is a classic trigger for trigonometric substitutions, specifically those involving sine or cosine. Why? Because we can leverage the Pythagorean identity sin^2(u) + cos^2(u) = 1 to simplify the expression under the radical.

Now, the question isn't just about finding any antiderivative; it's about understanding why a particular substitution, x = √5sin(u), is the preferred route and why alternatives might stumble. This is where the nuances of calculus, especially the importance of domain restrictions and the invertibility of trigonometric functions, come into play. We need to consider not just the algebraic manipulation but also the geometric implications of our choices.

So, what makes x = √5sin(u) so special? It's all about crafting a substitution that neatly transforms the integrand into a form we can easily integrate, while also ensuring that our substitution is invertible over a suitable domain. This invertibility is crucial because, after integrating with respect to 'u', we need to go back to our original variable, 'x'. If our substitution isn't invertible, we're in trouble! We'll explore this in greater depth as we unpack the trigonometric substitution technique.

Remember, guys, the goal here isn't just to memorize a formula or a trick. It's about developing a deep understanding of why certain methods work and others don't. This kind of conceptual grasp is what will truly empower you to tackle even the most challenging integrals. So, let's keep digging!

The Winning Move: x = √5sin(u) and the Trigonometric Tango

The substitution x = √5sin(u) is a stroke of genius in this case, and here's why. By making this move, we're not just swapping variables; we're setting off a chain reaction of simplification that leads us straight to the solution. Let's break down the magic step-by-step:

  1. The Substitution Itself: We let x = √5sin(u). This is the heart of the matter. But why sine? Because it plays perfectly with the (5 - x^2) term. Remember our Pythagorean identity friend, sin^2(u) + cos^2(u) = 1? We're about to put it to work.
  2. The Differential Dance: We need to find dx in terms of du. Differentiating both sides of x = √5sin(u), we get dx = √5cos(u) du. This is crucial for replacing dx in our original integral.
  3. The Transformation: Now, we substitute both x and dx into the integral: ∫(5 - x2)(-3/2) dx becomes ∫(5 - (√5sin(u))2)(-3/2) * √5cos(u) du. Brace yourselves, the simplification is coming!
  4. The Pythagorean Power: Let's simplify the expression inside the integral. (√5sin(u))^2 is simply 5sin^2(u). So, we have ∫(5 - 5sin2(u))(-3/2) * √5cos(u) du. Now, factor out a 5: ∫(5(1 - sin2(u)))(-3/2) * √5cos(u) du. And here's where the Pythagorean identity shines! 1 - sin^2(u) is just cos^2(u). Our integral now looks like this: ∫(5cos2(u))(-3/2) * √5cos(u) du.
  5. The Algebraic Acrobatics: Time to simplify the powers. (5cos2(u))(-3/2) can be rewritten as 5^(-3/2) * (cos2(u))(-3/2), which is 5^(-3/2) * cos^(-3)(u). Our integral is now: ∫5^(-3/2) * cos^(-3)(u) * √5cos(u) du. Notice that √5 is the same as 5^(1/2). So, we have 5^(-3/2) * 5^(1/2) which simplifies to 5^(-1). Our integral is now: ∫5^(-1) * cos^(-3)(u) * cos(u) du.
  6. The Cosine Cleanup: cos^(-3)(u) * cos(u) is cos^(-2)(u), which is 1/cos^2(u), or sec^2(u). So, our integral becomes: ∫(1/5) * sec^2(u) du.
  7. The Integration Illumination: The integral of sec^2(u) is tan(u)! So, we have (1/5)tan(u) + C, where C is the constant of integration.
  8. The Grand Finale: Back to x: We're not done yet! We need to express our answer in terms of x. This is where the invertibility of our substitution becomes crucial. We know x = √5sin(u), so sin(u) = x/√5. To find tan(u), we can think of a right triangle where the opposite side is x and the hypotenuse is √5. Using the Pythagorean theorem, the adjacent side is √(5 - x^2). Therefore, tan(u) = opposite/adjacent = x/√(5 - x^2).
  9. The Final Flourish: Substituting tan(u) back into our result, we get (1/5) * (x/√(5 - x^2)) + C, which simplifies to x / (5√(5 - x^2)) + C. This is our indefinite integral!

Phew! That was quite a journey, right? But it highlights the power of the x = √5sin(u) substitution. It transformed a seemingly complex integral into a manageable one. But what about other substitutions? Why might they fail us?

The Pitfalls of Alternative Paths: When Substitutions Go Wrong

Okay, so x = √5sin(u) is the hero of our story. But what if we tried a different substitution? Let's explore why some seemingly reasonable alternatives might lead us astray. This isn't just about finding the right answer; it's about understanding why certain approaches are flawed.

One common temptation might be to try a simple algebraic substitution, like u = 5 - x^2. This seems like it would simplify the expression inside the parentheses. Let's see what happens:

  1. The Algebraic Adventure: Let u = 5 - x^2.
  2. The Differential Detour: Differentiating both sides, we get du = -2x dx. So, dx = du / (-2x).
  3. The Substitution Stumble: Substituting into the integral, we get ∫(u)^(-3/2) * (du / (-2x)). Uh oh... we still have an 'x' in the integral! This is a major red flag. Our goal with substitution is to get rid of the original variable entirely. We need to express everything in terms of 'u'.
  4. The Dead End: We could try to solve u = 5 - x^2 for x, getting x = ±√(5 - u). But this introduces a square root and a ± sign, making the integral even messier. We're not simplifying things; we're making them worse!

Why did this happen? The substitution u = 5 - x^2 doesn't effectively deal with the square root in the denominator. It simplifies the expression inside the parentheses, but it doesn't eliminate the need for a trigonometric identity to handle the √(5 - x^2) term. We're essentially trading one problem for another, and the new problem is arguably more complex.

Another potential misstep could be trying a substitution like x = √5cos(u). This might seem similar to our successful substitution, but it can lead to complications if we're not careful about the domain of 'u'. While it will lead to a correct answer eventually, the algebraic manipulations are messier and the chance of making a mistake is higher.

The Key Lesson: Not all substitutions are created equal. The best substitutions are those that not only simplify the integrand but also allow us to easily express dx in terms of du and, crucially, allow us to invert the substitution to get back to our original variable, x. Trigonometric substitutions, particularly x = a sin(u) or x = a tan(u) for expressions of the form √(a^2 - x^2) or √(a^2 + x^2), are powerful tools because they leverage trigonometric identities to achieve this simplification and invertibility. However, you should also take extra care to correctly handle the inverse trigonometric functions, especially regarding their domain and range.

Domain Domination: Why Intervals Matter

Now, let's shift our focus to the interval of integration: [-1, 2]. Why is this interval significant, and how does it influence our choice of substitution and the interpretation of our result?

The interval [-1, 2] tells us the specific range of x-values we're interested in. This is crucial for two main reasons:

  1. Definite Integrals and Area: If we were calculating a definite integral (i.e., finding the area under the curve), the interval would define the limits of integration. We would need to change our limits of integration from x-values to u-values based on our substitution. This ensures we're calculating the area over the correct region.
  2. Domain Restrictions and Invertibility: The interval can also impact the validity of our substitution, particularly when dealing with trigonometric functions. Remember, trigonometric functions are periodic, meaning they repeat their values. This means that the inverse trigonometric functions (like arcsin, arccos, arctan) have restricted domains and ranges to ensure they are one-to-one (i.e., each input has a unique output).

In our case, with the substitution x = √5sin(u), we're implicitly using the arcsin function to go back from u to x. The arcsin function has a range of [-π/2, π/2]. This means that when we solve for u (u = arcsin(x/√5)), we need to make sure that the values of u we obtain fall within this range.

Let's consider the endpoints of our interval, [-1, 2]:

  • When x = -1, u = arcsin(-1/√5) ≈ -0.46 radians, which is within [-π/2, π/2].
  • When x = 2, u = arcsin(2/√5) ≈ 0.90 radians, which is also within [-π/2, π/2].

So, for this particular interval, the substitution x = √5sin(u) works seamlessly. The corresponding u-values fall within the principal range of arcsin, and we don't need to worry about any ambiguities.

But what if our interval was different? Suppose our interval was [2, 3]. When x = 3, u = arcsin(3/√5). But 3/√5 is greater than 1, and the arcsin function is only defined for values between -1 and 1! This means our substitution would not be valid over this interval.

The Takeaway: The interval of integration plays a critical role in determining the validity of our substitution. We need to ensure that the substitution is well-defined and invertible over the given interval, and that the resulting u-values fall within the appropriate ranges for the inverse trigonometric functions.

The Indefinite Integral Unveiled: A Recap and Final Thoughts

Alright guys, we've reached the end of our integral adventure! Let's recap the key insights we've gained:

  • The Power of Trigonometric Substitution: Integrals involving expressions of the form √(a^2 - x^2), √(a^2 + x^2), or √(x^2 - a^2) often yield beautifully to trigonometric substitutions. These substitutions leverage trigonometric identities to simplify the integrand and make it more manageable.
  • The Importance of Choosing Wisely: Not all substitutions are created equal. The best substitutions are those that simplify the integral, allow us to express dx in terms of du, and, crucially, are invertible so we can return to our original variable.
  • The Domain's Decree: The interval of integration matters! It influences the validity of our substitution and the interpretation of our results, especially when dealing with inverse trigonometric functions.
  • The Grand Finale: We successfully navigated the indefinite integral of (5 - x2)(-3/2) using the substitution x = √5sin(u), arriving at the solution x / (5√(5 - x^2)) + C.

But the most important takeaway is this: Mastering integration isn't just about memorizing techniques; it's about understanding why those techniques work. By delving into the nuances of substitution, domain restrictions, and trigonometric identities, you'll be well-equipped to tackle a wide range of integral challenges. Keep practicing, keep exploring, and never stop asking "why!"

So, next time you're faced with a tricky integral, remember our journey today. Think about the structure of the integrand, consider your substitution options carefully, and always keep the domain in mind. You've got this!