Antiderivative Of 1/(x√(x²-25)) Solution And Simplification

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Introduction

In this article, we will delve into the process of finding the antiderivative of the function 1/(x√(x²-25)). This type of integral often appears in calculus and requires a clever application of trigonometric substitution to solve effectively. We will systematically walk through the steps, providing detailed explanations to ensure a clear understanding of each stage. Our goal is not only to find the antiderivative but also to simplify the result to its most elegant form. This involves recognizing trigonometric identities and applying algebraic manipulations.

Understanding antiderivatives is crucial in calculus, as they are the foundation for solving many problems in physics, engineering, and other scientific fields. The antiderivative, also known as the indefinite integral, is the reverse operation of differentiation. When we find the antiderivative of a function, we are essentially finding a family of functions whose derivative is the original function. The constant of integration, denoted as 'C', accounts for the fact that the derivative of a constant is zero. Thus, there are infinitely many antiderivatives for a given function, differing only by a constant.

Trigonometric substitution is a powerful technique for evaluating integrals involving expressions of the form √(a² - x²), √(a² + x²), or √(x² - a²). The basic idea behind this method is to replace the variable 'x' with a trigonometric function, such as sine, tangent, or secant, to simplify the integral. The choice of trigonometric function depends on the form of the expression under the square root. For example, if we have √(a² - x²), we might use the substitution x = a sin(θ). If we have √(a² + x²), we might use x = a tan(θ). And if we have √(x² - a²), as in our case, we typically use x = a sec(θ). By making this substitution, we can often transform the integral into a simpler form that can be evaluated using standard integration techniques. After evaluating the integral in terms of θ, we need to convert back to the original variable 'x' using the relationship between x and θ.

Step-by-Step Solution

To find the antiderivative of the given function, ∫ 1/(x√(x²-25)) dx, we will employ trigonometric substitution. Here’s a detailed breakdown of the process:

1. Trigonometric Substitution

The integral we're tackling is ∫ 1/(x√(x²-25)) dx. Notice the term √(x² - 25) in the denominator. This suggests a trigonometric substitution using the secant function. Let's set:

x = 5 sec(θ)

This substitution is chosen because it aligns well with the form √(x² - a²), where a = 5 in our case. This substitution is a cornerstone of solving integrals involving such expressions, effectively transforming the algebraic challenge into a trigonometric one. The magic of this substitution lies in its ability to simplify the square root term, a crucial step in making the integral more manageable. The constant 5 is strategically chosen to match the constant within the square root, ensuring that the trigonometric identity we'll use later fits perfectly.

Now, we need to find dx in terms of dθ. Differentiating both sides of x = 5 sec(θ) with respect to θ, we get:

dx = 5 sec(θ) tan(θ) dθ

This derivative is a crucial link between the original variable x and the new variable θ. It allows us to replace dx in the original integral, completing the transformation from the x-domain to the θ-domain. The derivative of sec(θ) is a standard result in calculus, and understanding its derivation is key to mastering trigonometric substitution techniques. The resulting expression for dx in terms of dθ is a bit more complex than a simple constant multiple, reflecting the non-linear nature of the secant function.

2. Substitute into the Integral

Now, we substitute x and dx into the original integral:

∫ 1/(x√(x²-25)) dx = ∫ 1/(5 sec(θ)√((5 sec(θ))² - 25)) * 5 sec(θ) tan(θ) dθ

This substitution is the heart of the method, where we replace every instance of x and dx in the original integral with their equivalents in terms of θ. This step transforms the integral into a new form that, while initially appearing more complex, is actually structured to leverage trigonometric identities for simplification. The care taken in choosing the substitution now pays off, as we've set the stage for the square root to simplify nicely. The presence of both sec(θ) and tan(θ) in the substituted integral hints at the trigonometric identity we'll be using next, connecting the squares of these functions.

3. Simplify the Expression

First, simplify the square root term:

√((5 sec(θ))² - 25) = √(25 sec²(θ) - 25) = √(25(sec²(θ) - 1))

This simplification is a crucial step, leveraging algebraic manipulation to prepare the expression for the application of a trigonometric identity. Factoring out the 25 is a key move, revealing the familiar expression (sec²(θ) - 1) inside the square root. This manipulation highlights the strategic nature of trigonometric substitution, where the initial choice of substitution is designed to create opportunities for such simplifications. The goal here is to make the expression under the square root a perfect square, which will allow us to eliminate the square root in the next step.

Using the trigonometric identity sec²(θ) - 1 = tan²(θ):

√(25(sec²(θ) - 1)) = √(25 tan²(θ)) = 5 |tan(θ)|

This step is where the trigonometric identity comes into play, transforming the expression under the square root into a simpler form involving the tangent function. The identity sec²(θ) - 1 = tan²(θ) is a cornerstone of trigonometry and is essential for solving integrals using trigonometric substitution. The absolute value signs around tan(θ) are a reminder that the square root of a square is the absolute value. However, for simplicity, and assuming we are working in an interval where tan(θ) is positive, we can often drop the absolute value signs in subsequent steps, but it's important to be mindful of this detail.

Now, substitute this back into the integral:

∫ 1/(5 sec(θ) * 5 |tan(θ)|) * 5 sec(θ) tan(θ) dθ

This substitution brings us closer to simplifying the integral. We've replaced the complex square root term with a simpler expression involving tan(θ). The presence of sec(θ) and tan(θ) both in the numerator and denominator suggests that further cancellations are possible, a hallmark of successful trigonometric substitution. The overall strategy here is to transform the integral into a form where we can apply basic integration rules, and this substitution is a significant step in that direction.

4. Cancel Terms and Integrate

We can cancel out the common terms:

∫ (5 sec(θ) tan(θ)) / (5 sec(θ) * 5 |tan(θ)|) dθ = ∫ tan(θ) / (5 |tan(θ)|) dθ

This cancellation step is a direct result of the previous substitutions and simplifications. It demonstrates the power of trigonometric substitution in transforming a complex integral into a much simpler one. The sec(θ) terms cancel out neatly, and we're left with an expression involving only tan(θ). The absolute value signs around tan(θ) are still present, but we'll address them in the next step. The goal here is to isolate a basic trigonometric function that we can integrate directly.

Assuming tan(θ) is positive (which is a common assumption in these problems for simplicity):

∫ tan(θ) / (5 |tan(θ)|) dθ = ∫ 1/5 dθ

This simplification is a crucial step that allows us to integrate the function directly. By assuming tan(θ) is positive, we can drop the absolute value signs and cancel tan(θ) from the numerator and denominator. This assumption is often made in the context of these types of problems to simplify the calculations, but it's important to remember that the general solution might require considering cases where tan(θ) is negative as well. The resulting integral is now a simple constant multiple of dθ, which is straightforward to integrate.

Now, integrate with respect to θ:

∫ 1/5 dθ = (1/5)θ + C

This integration step is the culmination of the simplification process, where we finally find the antiderivative with respect to θ. The integral of a constant is simply the constant times the variable of integration, in this case, θ. The constant of integration, C, is added to account for the fact that the derivative of a constant is zero, meaning there are infinitely many possible antiderivatives that differ only by a constant. This constant of integration is a crucial part of the indefinite integral, representing the family of functions that have the same derivative.

5. Convert Back to x

Finally, we need to convert back to the original variable x. Recall that we made the substitution:

x = 5 sec(θ)

From this, we can find θ:

sec(θ) = x/5 θ = arcsec(x/5)

This conversion back to the original variable x is a critical step in solving indefinite integrals using substitution techniques. We need to express our result in terms of the original variable, as the problem was posed in terms of x. The relationship between θ and x is derived directly from our initial substitution. The arcsecant function, also known as the inverse secant function, is the inverse of the secant function, and it allows us to solve for θ in terms of x. Understanding inverse trigonometric functions is essential for completing the trigonometric substitution method.

Substitute θ back into our result:

(1/5)θ + C = (1/5)arcsec(x/5) + C

This final substitution completes the process of finding the antiderivative. We've replaced θ with its expression in terms of x, giving us the antiderivative in terms of the original variable. The constant of integration, C, remains, as it's an essential part of the indefinite integral. The function arcsec(x/5) is the inverse secant function, and it represents the angle whose secant is x/5. The result (1/5)arcsec(x/5) + C is the general antiderivative of the original function, meaning that its derivative is equal to 1/(x√(x²-25)).

Final Answer

Therefore, the antiderivative of 1/(x√(x²-25)) is:

∫ 1/(x√(x²-25)) dx = (1/5)arcsec(x/5) + C

This final answer represents the indefinite integral of the original function. It's a family of functions that all have the same derivative, differing only by a constant. The arcsecant function is a key part of the solution, reflecting the use of trigonometric substitution in solving the integral. The constant of integration, C, is included to emphasize that there are infinitely many possible antiderivatives. This solution demonstrates the power and elegance of trigonometric substitution as a technique for solving integrals that might otherwise be intractable.

Conclusion

In this article, we successfully found the antiderivative of 1/(x√(x²-25)) using trigonometric substitution. We detailed each step, from choosing the appropriate substitution to simplifying the integral and converting back to the original variable. The key steps involved substituting x = 5 sec(θ), simplifying the resulting expression using trigonometric identities, integrating with respect to θ, and finally converting back to x. This example illustrates the importance of trigonometric substitution in solving integrals involving square roots of quadratic expressions.

Trigonometric substitution is a versatile technique that can be applied to a wide range of integrals. Mastering this method requires a solid understanding of trigonometric identities and the derivatives and integrals of trigonometric functions. The process involves carefully choosing a substitution that simplifies the integral, performing the substitution, simplifying the resulting expression, integrating, and then converting back to the original variable. This technique is a powerful tool in the calculus toolbox, allowing us to solve integrals that would be difficult or impossible to solve using other methods.

The ability to find antiderivatives is fundamental to calculus and has numerous applications in various fields. Whether it's calculating areas under curves, solving differential equations, or modeling physical phenomena, the concept of the antiderivative is essential. By understanding and mastering techniques like trigonometric substitution, we can tackle a wider range of problems and gain a deeper appreciation for the power of calculus. The steps outlined in this article provide a clear roadmap for approaching similar integrals, empowering readers to solve complex problems with confidence. The combination of algebraic manipulation, trigonometric identities, and careful substitution is a hallmark of successful integration, and this example serves as a valuable illustration of these principles in action.