Solving ∫ 75 (|x+2| √((x+2)² - 25)) Dx A Step-by-Step Guide
Hey there, math enthusiasts! Ever stumbled upon an integral that looks like it belongs in a superhero movie? Something with intense symbols and a vibe that screams, "Solve me if you dare!"? Well, today, we're diving headfirst into one such integral. We're going to break down the fascinating integral of ∫ 75 / (|x+2| √((x+2)² - 25)) dx, explore its twists and turns, and emerge victorious with the solution. So, buckle up, grab your mathematical swords, and let's get started!
Setting the Stage: The Integral in Question
Before we jump into the nitty-gritty, let's take a good look at our star player: ∫ 75 / (|x+2| √((x+2)² - 25)) dx. At first glance, it might seem intimidating, but don't worry, we're going to dissect it piece by piece. This integral falls into the category of integrals that can be solved using trigonometric substitution, a powerful technique that transforms algebraic expressions into trigonometric ones, making them easier to handle. The presence of the square root term, √((x+2)² - 25), is a major clue that trigonometric substitution is our friend here. Specifically, the form √(u² - a²) suggests a secant substitution, which we'll explore in detail shortly. Remember, integrals like this aren't just abstract math; they pop up in various fields, from physics to engineering, modeling everything from the motion of objects to the flow of electricity. Understanding how to solve them is a valuable skill in any STEM field. So, let's roll up our sleeves and get ready to tackle this mathematical beast!
Trigonometric Substitution: Our Heroic Transformation
Okay, guys, here's where the magic happens! Trigonometric substitution is our secret weapon for tackling integrals with square roots of quadratic expressions. In our case, the integral ∫ 75 / (|x+2| √((x+2)² - 25)) dx has a term that looks suspiciously like √(u² - a²), where 'u' is (x+2) and 'a' is 5. This form screams for a secant substitution. Why secant? Well, recall the trigonometric identity sec²θ - 1 = tan²θ. If we let (x+2) = 5 secθ, then (x+2)² - 25 becomes 25 sec²θ - 25, which simplifies to 25(sec²θ - 1) = 25 tan²θ. Taking the square root gives us 5 tanθ, a much simpler expression to deal with. So, let's make the substitution: x + 2 = 5 secθ. This is our pivotal move, the key to unlocking the integral's secrets. Now, we need to find dx in terms of dθ. Differentiating both sides with respect to θ, we get dx = 5 secθ tanθ dθ. This is another crucial piece of the puzzle. With our substitution and the expression for dx, we're ready to transform the integral into a trigonometric wonderland. Stay tuned, because the next step is where we'll see the integral shed its algebraic skin and reveal its trigonometric core. This substitution isn't just a trick; it's a fundamental technique in calculus, allowing us to simplify complex expressions and make integration possible. Mastering this technique opens doors to solving a wide range of integrals that initially seem daunting.
Substituting and Simplifying: Taming the Trigonometric Beast
Alright, mathletes, let's put our substitution to work and simplify the heck out of this integral! Remember, we've got x + 2 = 5 secθ and dx = 5 secθ tanθ dθ. Now, we're going to plug these into our original integral: ∫ 75 / (|x+2| √((x+2)² - 25)) dx. First, let's replace (x+2) with 5 secθ and dx with 5 secθ tanθ dθ. The integral now looks like this: ∫ 75 / (|5 secθ| √((5 secθ)² - 25)) * 5 secθ tanθ dθ. Notice that absolute value! It's super important because the sign of secθ depends on the range of θ, and we need to handle it carefully. Now, let's simplify the square root: √((5 secθ)² - 25) = √(25 sec²θ - 25) = √(25(sec²θ - 1)) = √(25 tan²θ) = 5 |tanθ|. Again, the absolute value is crucial. Our integral now looks even more exciting: ∫ 75 / (|5 secθ| * 5 |tanθ|) * 5 secθ tanθ dθ. Time for some serious cancellation! We can pull the constant 75 outside the integral: 75 ∫ (5 secθ tanθ) / (|5 secθ| * 5 |tanθ|) dθ. Now, let's simplify the constants: 75 ∫ (secθ tanθ) / (|secθ| * |tanθ|) dθ. This is where things get interesting. The absolute values |secθ| and |tanθ| require us to consider different cases based on the signs of secθ and tanθ. We'll tackle those cases in the next section, but for now, we've made significant progress in transforming our integral into a more manageable form. The key takeaway here is the power of substitution in simplifying complex integrals. By carefully choosing our substitution and applying trigonometric identities, we've moved closer to the solution. This step-by-step approach is crucial for tackling any challenging integral, so let's keep pushing forward!
Cases and Considerations: Navigating the Absolute Values
Okay, team, we've arrived at the tricky part where we need to deal with those pesky absolute values! Remember our integral: 75 ∫ (secθ tanθ) / (|secθ| * |tanθ|) dθ. The absolute values |secθ| and |tanθ| mean we need to consider different cases based on the signs of secθ and tanθ. Let's break it down:
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Case 1: secθ > 0 and tanθ > 0
In this case, |secθ| = secθ and |tanθ| = tanθ. Our integral simplifies beautifully: 75 ∫ (secθ tanθ) / (secθ tanθ) dθ = 75 ∫ 1 dθ. This is a simple integral that we can easily solve.
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Case 2: secθ < 0 and tanθ < 0
Here, |secθ| = -secθ and |tanθ| = -tanθ. The integral becomes: 75 ∫ (secθ tanθ) / ((-secθ) * (-tanθ)) dθ = 75 ∫ 1 dθ. Surprise! We get the same simplified integral as in Case 1.
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Case 3: secθ > 0 and tanθ < 0
Now, |secθ| = secθ and |tanθ| = -tanθ. Our integral transforms into: 75 ∫ (secθ tanθ) / (secθ * (-tanθ)) dθ = 75 ∫ -1 dθ.
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Case 4: secθ < 0 and tanθ > 0
Finally, |secθ| = -secθ and |tanθ| = tanθ. The integral becomes: 75 ∫ (secθ tanθ) / ((-secθ) * tanθ) dθ = 75 ∫ -1 dθ. So, we have two possible simplified integrals: 75 ∫ 1 dθ and 75 ∫ -1 dθ. This might seem a bit confusing, but it highlights the importance of carefully considering absolute values. The different cases arise from the fact that the original integral involves |x+2|, which can be positive or negative depending on the value of x. We've essentially split the problem into different intervals where the expression inside the absolute value has a consistent sign. Now that we've handled the absolute values, we're ready to integrate and find our solution. Remember, this casework is a crucial step in solving integrals involving absolute values, ensuring we account for all possible scenarios.
Integrating and Back-Substituting: The Final Stretch
Alright, everyone, we're in the home stretch! We've navigated the trigonometric substitution, tamed the absolute values, and now it's time to integrate and bring it all back home. From the previous section, we have two possible integrals:
- 75 ∫ 1 dθ = 75θ + C₁
- 75 ∫ -1 dθ = -75θ + C₂
Where C₁ and C₂ are constants of integration. Now, we need to back-substitute to get our answer in terms of x. Remember our original substitution: x + 2 = 5 secθ. To solve for θ, we take the inverse secant (arcsec): θ = arcsec((x+2)/5). This is the crucial link that connects our trigonometric world back to the original algebraic one. Let's substitute this back into our integrated expressions:
- 75θ + C₁ = 75 arcsec((x+2)/5) + C₁
- -75θ + C₂ = -75 arcsec((x+2)/5) + C₂
These are our two possible solutions, depending on the cases we considered earlier. However, we can simplify this further. Notice that the original integral involves |x+2|, which means the solution should also reflect this symmetry. We can combine the two cases into a single solution by using the properties of the arcsecant function and absolute values. The general solution is: ∫ 75 / (|x+2| √((x+2)² - 25)) dx = 15 arcsec(|(x+2)/5|) + C. Here, C is the constant of integration, which absorbs both C₁ and C₂. We've done it! We've successfully integrated the original expression and found the solution. This final step of back-substitution is essential, as it transforms our answer from the trigonometric domain back to the original variable, x. The arcsecant function arises naturally from our secant substitution, highlighting the power of choosing the right substitution technique. And remember, the constant of integration, C, is always added to indefinite integrals, representing the family of functions that have the same derivative.
Conclusion: Victory Lap!
Wow, guys, we've been on quite the mathematical journey! We started with a seemingly intimidating integral, ∫ 75 / (|x+2| √((x+2)² - 25)) dx, and, step by step, we conquered it. We employed the powerful technique of trigonometric substitution, specifically a secant substitution, to transform the integral into a more manageable form. We carefully navigated the complexities of absolute values, considering different cases based on the signs of trigonometric functions. We integrated the simplified expressions and then, crucially, back-substituted to express our solution in terms of the original variable, x. Our final answer is: 15 arcsec(|(x+2)/5|) + C. This journey highlights the beauty and power of calculus. It demonstrates how seemingly complex problems can be broken down into smaller, more manageable steps. It showcases the importance of choosing the right techniques, like trigonometric substitution, and the necessity of careful consideration, especially when dealing with absolute values. But most importantly, it underscores the satisfaction of arriving at a solution after a challenging mathematical adventure. So, the next time you encounter a daunting integral, remember this journey. Remember the power of trigonometric substitution, the importance of casework, and the joy of mathematical discovery. You've got this!
