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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 integration technique. It's not a simple power rule integration, and direct substitution won't suffice. This comprehensive guide will explore various approaches to solve this integral, delving into the intricacies of trigonometric substitution and hyperbolic substitution, and providing a detailed understanding of the solution and its applications.

Understanding the Problem: Why Standard Techniques Fail

Before diving into the solution, let's understand why standard integration techniques are insufficient. The presence of the square root makes simple u-substitution ineffective. Integration by parts also won't readily yield a solution. The key to solving this integral lies in recognizing the structure of the integrand and employing a strategic substitution that simplifies the expression.

Method 1: Trigonometric Substitution

This is arguably the most common and intuitive method for solving this integral. The key is to recognize the Pythagorean identity: sin²θ + cos²θ = 1. We can manipulate this identity to resemble the expression under the square root.

1. The Substitution:

We make the substitution: x = a tan θ. This implies dx = a sec²θ dθ.

2. Substituting into the Integral:

Substituting x and dx into the original integral, we get:

∫√( (a tan θ)² + a²) * a sec²θ dθ

3. Simplifying the Integrand:

This simplifies to:

∫√(a²(tan²θ + 1)) * a sec²θ dθ = ∫√(a²sec²θ) * a sec²θ dθ = ∫a sec θ * a sec²θ dθ = a² ∫sec³θ dθ

4. Solving the Integral of sec³θ:

This is a well-known integral that requires integration by parts. Let's break it down:

• Let u = sec θ and dv = sec²θ dθ
• Then du = sec θ tan θ dθ and v = tan θ

Applying integration by parts: ∫ u dv = uv - ∫ v du, we get:

a² ∫sec³θ dθ = a² [sec θ tan θ - ∫ tan²θ sec θ dθ]

Since tan²θ = sec²θ - 1, we can further simplify:

a² [sec θ tan θ - ∫ (sec²θ - 1) sec θ dθ] = a² [sec θ tan θ - ∫ sec³θ dθ + ∫ sec θ dθ]

Notice that the integral ∫sec³θ dθ appears on both sides. Solving for this integral:

2a² ∫sec³θ dθ = a² [sec θ tan θ + ln|sec θ + tan θ|]

Therefore:

∫sec³θ dθ = (a²/2) [sec θ tan θ + ln|sec θ + tan θ|]

5. Back-substitution:

Now, we need to substitute back in terms of x. Recall that x = a tan θ, which means tan θ = x/a. We can construct a right-angled triangle with opposite side x and adjacent side a, making the hypotenuse √(x² + a²). Therefore:

sec θ = √(x² + a²)/a

Substituting these values back into the solution:

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

6. Final Result (Trigonometric Substitution):

Simplifying the expression, we arrive at the final result:

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

Method 2: Hyperbolic Substitution

This method provides an alternative approach, often considered more elegant by some. It leverages the hyperbolic identities.

1. The Substitution:

We substitute x = a sinh u. This implies dx = a cosh u du.

2. Substituting into the Integral:

Substituting x and dx into the original integral, we obtain:

∫√((a sinh u)² + a²) * a cosh u du

3. Simplifying the Integrand:

Using the hyperbolic identity cosh²u - sinh²u = 1, we simplify the expression:

∫√(a²(sinh²u + 1)) * a cosh u du = ∫√(a²cosh²u) * a cosh u du = ∫a²cosh²u du

4. Solving the Integral of cosh²u:

We utilize the identity cosh²u = (cosh 2u + 1)/2:

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

5. Back-substitution:

Recall that x = a sinh u, so sinh u = x/a. Using the identity sinh 2u = 2 sinh u cosh u = 2 sinh u √(1 + sinh²u), we can express sinh 2u and u in terms of x:

sinh 2u = 2(x/a)√(1 + (x/a)²) = (2x/a²)√(a² + x²)

u = sinh⁻¹(x/a) = ln|x + √(x² + a²)| - ln|a|

6. Final Result (Hyperbolic Substitution):

Substituting back and simplifying:

(a²/2) [(1/2)(2x/a²)√(x² + a²) + ln|x + √(x² + a²)| - ln|a|] + C = (x/2)√(x² + a²) + (a²/2)ln|x + √(x² + a²)| + C'

Where C' incorporates the constant term from the ln|a|. Note that this result is identical to the result obtained through trigonometric substitution, except for the constant of integration.

Applications of the Integral

This integral appears in various applications across different fields:

• Calculating Arc Length: Determining the arc length of curves, particularly those defined by functions involving quadratic terms.
• Surface Area Calculations: Calculating the surface area of revolution when rotating curves around an axis.
• Physics: Solving problems involving motion under gravity or related to electrical fields.
• Engineering: Various engineering applications require this integral to solve problems related to stress, strain, and other physical quantities.

Conclusion

The integral ∫√(x² + a²) dx, while seemingly straightforward, demands a specific integration technique. Both trigonometric and hyperbolic substitution methods provide valid and equivalent solutions. The choice between them often comes down to personal preference and the context of the problem. Understanding both methods is crucial for a comprehensive grasp of this integral and its broader applications in diverse fields. Remember to always check your work and consider the different techniques to find the most efficient and elegant solution. This guide hopefully provided a thorough and detailed exploration of this fundamental integral.