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Integrating ∫√(x² + a²) dx: A Comprehensive Guide

The integral ∫√(x² + a²) dx is a classic example of an integral that requires a specific technique to solve. It's not a straightforward power rule integration, and often stumps students new to calculus. This comprehensive guide will walk you through various methods for solving this integral, explaining the underlying principles and providing examples to solidify your understanding. We'll cover trigonometric substitution, hyperbolic substitution, and discuss the application of these methods in different scenarios.

Understanding the Challenge

The integral ∫√(x² + a²) dx is challenging because the square root of a sum of squares doesn't have a simple antiderivative. Unlike ∫√(x² - a²) dx or ∫√(a² - x²) dx, which utilize trigonometric substitutions based on Pythagorean identities directly, this integral requires a slightly different approach.

Method 1: Trigonometric Substitution

This is the most common method for solving this type of integral. The key is to use a trigonometric substitution that transforms the expression x² + a² into a form that can be simplified using trigonometric identities.

1.1 The Substitution:

We use the substitution x = a tan θ. This implies dx = a sec² θ dθ.

1.2 Substituting into the Integral:

Substituting x = a tan θ and dx = a sec² θ dθ into the integral ∫√(x² + a²) dx, we get:

∫√((a tan θ)² + a²) * a sec² θ dθ = ∫√(a²(tan² θ + 1)) * a sec² θ dθ

1.3 Simplifying using Trigonometric Identities:

Recall the Pythagorean identity: tan² θ + 1 = sec² θ. Substituting this into the integral simplifies it considerably:

∫√(a² sec² θ) * a sec² θ dθ = ∫a sec θ * a sec² θ dθ = a² ∫sec³ θ dθ

1.4 Solving the Integral of sec³ θ:

The integral of sec³ θ is a standard integral that can be solved using integration by parts. The process is as follows:

Let u = sec θ and dv = sec² θ dθ. Then du = sec θ tan θ dθ and v = tan θ.

Using the integration by parts formula (∫u dv = uv - ∫v du), we get:

∫sec³ θ dθ = sec θ tan θ - ∫tan² θ sec θ dθ

Using the identity tan² θ = sec² θ - 1, we can further simplify:

∫sec³ θ dθ = sec θ tan θ - ∫(sec² θ - 1) sec θ dθ = sec θ tan θ - ∫sec³ θ dθ + ∫sec θ dθ

Now, we can solve for ∫sec³ θ dθ:

2∫sec³ θ dθ = sec θ tan θ + ∫sec θ dθ = sec θ tan θ + ln|sec θ + tan θ| + C

Therefore, ∫sec³ θ dθ = (1/2) [sec θ tan θ + ln|sec θ + tan θ|] + C

1.5 Back-substitution:

Substitute back into the original integral:

a² ∫sec³ θ dθ = (a²/2) [sec θ tan θ + ln|sec θ + tan θ|] + C

Since x = a tan θ, we have tan θ = x/a. Also, sec θ = √(1 + tan² θ) = √(1 + (x/a)²) = √(x² + a²)/a.

Substituting these back into the expression, we finally get:

(a²/2) [(x/a)√(x² + a²)/a + ln| (√(x² + a²)/a) + (x/a) |] + C

Simplifying, the final answer is:

(1/2) [x√(x² + a²) + a² ln|x + √(x² + a²)|] + C

Method 2: Hyperbolic Substitution

Another method to solve ∫√(x² + a²) dx is using hyperbolic substitution. This method offers a slightly different approach and can be preferable in certain contexts.

2.1 The Substitution:

We substitute x = a sinh u, where sinh u is the hyperbolic sine function. This implies dx = a cosh u du.

2.2 Substituting into the Integral:

Substituting x = a sinh u and dx = a cosh u du into the integral, we get:

∫√((a sinh u)² + a²) * a cosh u du = ∫√(a²(sinh² u + 1)) * a cosh u du

2.3 Simplifying using Hyperbolic Identities:

Recall the hyperbolic identity: cosh² u - sinh² u = 1, which implies 1 + sinh² u = cosh² u. Substituting this simplifies the integral:

∫√(a² cosh² u) * a cosh u du = ∫a cosh u * a cosh u du = a² ∫cosh² u du

2.4 Solving the Integral of cosh² u:

The integral of cosh² u can be solved using the identity cosh² u = (cosh 2u + 1)/2:

a² ∫(cosh 2u + 1)/2 du = (a²/2) ∫(cosh 2u + 1) du = (a²/2) [(sinh 2u)/2 + u] + C

2.5 Back-substitution:

Using the identity sinh 2u = 2 sinh u cosh u, and since x = a sinh u, we have sinh u = x/a. Also, cosh u = √(1 + sinh² u) = √(1 + (x/a)²) = √(x² + a²)/a.

Substituting back, we get:

(a²/2) [(2(x/a)(√(x² + a²)/a))/2 + sinh⁻¹(x/a)] + C = (1/2)[x√(x² + a²) + a²sinh⁻¹(x/a)] + C

Since sinh⁻¹(x/a) = ln|x + √(x² + a²)| - ln|a|, the final result, after simplification, is consistent with the trigonometric substitution method:

(1/2) [x√(x² + a²) + a² ln|x + √(x² + a²)|] + C

Comparing Methods

Both trigonometric and hyperbolic substitutions yield the same result. The choice between them often depends on personal preference and the context of the problem. Hyperbolic substitutions might be slightly simpler in certain cases, but the trigonometric method is generally more widely taught and understood.

Applications and Extensions

The integral ∫√(x² + a²) dx has numerous applications in various fields, including:

• Physics: Calculating arc lengths, work done against gravity, and other physical quantities often involve integrals of this form.
• Engineering: Similar to physics, many engineering applications, especially those dealing with curves and surfaces, use this integral.
• Geometry: Finding the area under certain curves or the volume of solids of revolution can require this integration technique.

This integral can be extended to more complex scenarios. For instance, integrals of the form ∫√(ax² + bx + c) dx can often be manipulated into a form solvable using the techniques outlined above through completing the square or other algebraic manipulations.

Conclusion

Mastering the integration of ∫√(x² + a²) dx is crucial for any student of calculus. This article provided a detailed explanation of two effective methods: trigonometric substitution and hyperbolic substitution. Understanding both methods allows for flexibility in approaching similar integrals. Remember that practice is key to mastering these techniques. By working through various examples and applying the principles explained above, you can build a solid understanding of how to approach and solve this important class of integrals. Always double-check your work and ensure you understand the underlying reasoning behind each step of the solution process. This will not only enhance your problem-solving skills but also deepen your overall understanding of calculus.