Solving ∫ 6 / |2x-1|√((2x-1)²-1) Dx A Step-by-Step Guide
Hey guys! Today, we're diving deep into a fascinating integral problem that might seem a bit intimidating at first glance. But don't worry, we'll break it down step-by-step, making sure you grasp every concept along the way. Our mission? To solve the integral: ∫ 6 / |2x-1|√((2x-1)²-1) dx. Buckle up, because we're about to embark on a mathematical adventure!
Understanding the Integral and Recognizing the Pattern
Okay, let's get started! When you first look at the integral ∫ 6 / |2x-1|√((2x-1)²-1) dx, it might seem a bit complex, but there's a hidden pattern we can exploit. The key here is to recognize the structure within the integral. Specifically, the expression inside the square root, (2x-1)²-1, and the term |2x-1| in the denominator should ring a bell. This structure closely resembles the derivative of the arcsecant function. Remember, the derivative of arcsec(u) is given by du / (|u|√(u²-1)). This is a crucial observation that guides our solution strategy.
Now, why is recognizing this pattern so important? Well, it allows us to make a strategic substitution that simplifies the integral significantly. By recognizing the connection to the arcsecant function, we can transform the integral into a more manageable form. Think of it as finding the right key to unlock a mathematical puzzle! This step is where a solid understanding of trigonometric derivatives and their corresponding integral forms really shines. We're essentially reverse-engineering the differentiation process to find the antiderivative.
To further clarify, let's delve into the arcsecant function a bit more. The arcsecant, often written as arcsec(x) or sec⁻¹(x), is the inverse function of the secant function. It essentially answers the question: "What angle has a secant equal to x?" Understanding its derivative is paramount for solving integrals of this type. The derivative of arcsec(x) is 1 / (|x|√(x²-1)). The presence of the absolute value and the square root term is what makes this particular form stand out and what we're leveraging in our integral problem. By identifying this characteristic form within the integral, we're setting ourselves up for a smooth and efficient solution.
The U-Substitution Technique: A Step-by-Step Approach
Alright, now that we've identified the arcsecant pattern, it's time to put our knowledge into action! The perfect tool for this job is the u-substitution technique. This method allows us to simplify integrals by replacing a complex expression with a single variable, making the integration process much cleaner and more straightforward.
In our case, the obvious choice for 'u' is 2x-1. Let's break this down step-by-step:
- Define u: Let u = 2x - 1. This substitution targets the expression inside the square root and the absolute value, which are the core components of the arcsecant form.
- Find du: Differentiate both sides of the equation with respect to x. So, du/dx = 2, which means du = 2 dx. We need to isolate dx, so we rewrite this as dx = du/2. This step is crucial for replacing the dx term in the original integral with an expression involving du.
- Substitute into the integral: Now, we replace every instance of 2x-1 with 'u' and dx with du/2 in the original integral. Our integral ∫ 6 / |2x-1|√((2x-1)²-1) dx transforms into ∫ 6 / |u|√(u²-1) (du/2). See how things are starting to look simpler?
- Simplify the integral: We can pull the constant factors out of the integral. The 6 in the numerator and the 2 in the denominator simplify to 3. This gives us 3 ∫ 1 / |u|√(u²-1) du. Now, doesn't that look much more familiar and less daunting?
This substitution process is like performing a magic trick on the integral. By carefully choosing our 'u' and making the appropriate substitutions, we've transformed a seemingly complex integral into a standard form that we readily recognize. The next step is to apply our knowledge of arcsecant integrals to finish the job.
Integrating and Back-Substituting: Bringing It All Together
Great job, guys! We've successfully simplified the integral using u-substitution. Now comes the exciting part: actually integrating and getting to our final solution. Remember, we've transformed our integral into 3 ∫ 1 / |u|√(u²-1) du. The form of this integral should immediately trigger our knowledge of arcsecant integrals. We know that the integral of 1 / (|u|√(u²-1)) is arcsec(u) + C, where C is the constant of integration. This is a fundamental integral that's essential to have in your mathematical toolkit.
Applying this knowledge to our transformed integral, we get: 3 ∫ 1 / |u|√(u²-1) du = 3 arcsec(u) + C. We're almost there, but we're not quite done yet! Remember, our original integral was in terms of x, so we need to substitute back to express our solution in terms of x as well. This is a critical step, as leaving the solution in terms of 'u' would be incomplete. It's like building a bridge and forgetting to connect it to the other side!
Recall that we defined u as 2x - 1. So, to back-substitute, we replace 'u' with 2x - 1 in our integrated expression. This gives us 3 arcsec(2x - 1) + C. And there you have it! We've successfully integrated the original expression and found the antiderivative.
This final step of back-substitution is a reminder that our ultimate goal is to solve the integral in terms of the original variable. It's a crucial part of the process and ensures that our solution is complete and meaningful. By carefully substituting back, we've linked our solution back to the original problem, providing a clear and accurate answer.
The Final Answer and Conclusion
Drumroll, please! After our mathematical journey through u-substitution and arcsecant integrals, we've arrived at the final answer. The integral ∫ 6 / |2x-1|√((2x-1)²-1) dx evaluates to:
3 arcsec(2x - 1) + C
Comparing this to the given options, we see that this corresponds to option a) 2 arcsec(2x-1) + C.
Option a states 2 arcsec(2x-1) + C, which is very close to our calculated answer. However, there seems to be a slight discrepancy in the coefficient. Our calculated coefficient is 3, whereas option a suggests 2. It's always good to double-check our work to ensure accuracy. Let's quickly review the steps:
- U-Substitution: u = 2x - 1, du = 2 dx, dx = du/2
- Transformed Integral: ∫ 6 / |u|√(u²-1) (du/2) = 3 ∫ 1 / |u|√(u²-1) du
- Integration: 3 ∫ 1 / |u|√(u²-1) du = 3 arcsec(u) + C
- Back-Substitution: 3 arcsec(2x - 1) + C
After reviewing, it seems our calculation is correct. The integral indeed evaluates to 3 arcsec(2x - 1) + C. Therefore, the closest option is not exactly matching, there might be a typo in the options provided or further simplification is possible which is not apparent right away.
So, there you have it, folks! We've successfully tackled a challenging integral by recognizing the arcsecant pattern, employing the u-substitution technique, and carefully integrating and back-substituting. Remember, the key to mastering integrals is to practice, recognize patterns, and don't be afraid to break down complex problems into smaller, more manageable steps. Keep up the great work, and happy integrating!
