Evaluate The Integral. 9 0 Dt 81 + T2

The integral ∫ 9 / (81 + t²) dt from 0 to ∞ represents the area under the curve of the function 9 / (81 + t²) between the limits 0 and infinity. Evaluating this integral requires a strong understanding of calculus, particularly techniques for integrating rational functions involving quadratic terms. This article will guide you through the step-by-step process of evaluating this integral, exploring the relevant mathematical concepts and providing insights into the underlying theory.

Understanding the Integral

Before diving into the solution, let's understand what we are trying to achieve. The integral ∫ 9 / (81 + t²) dt signifies finding the antiderivative of the function 9 / (81 + t²) and then evaluating it at the limits of integration, which are 0 and infinity. The antiderivative, also known as the indefinite integral, represents a function whose derivative is equal to the given function. In this case, we want to find a function whose derivative with respect to t is 9 / (81 + t²).

The presence of the term "81 + t²" in the denominator suggests that we might be able to use a trigonometric substitution to simplify the integral. Specifically, we can leverage the fact that the derivative of the arctangent function involves a similar form in the denominator.

Steps to Evaluate the Integral

Here's a detailed breakdown of the steps involved in evaluating the integral ∫ 9 / (81 + t²) dt from 0 to ∞:

1. Simplify the Integral:

The first step is to simplify the integral by factoring out any constants. In this case, we can factor out the constant 9 from the integral:

∫ 9 / (81 + t²) dt = 9 ∫ 1 / (81 + t²) dt

This simplifies the integral and makes it easier to work with. Now we need to focus on evaluating the integral ∫ 1 / (81 + t²) dt.

2. Trigonometric Substitution:

The form of the denominator, a² + t², where a² = 81, suggests a trigonometric substitution. We'll use the substitution:

t = a tan(θ) = 9 tan(θ)

This implies that:

dt = 9 sec²(θ) dθ

This substitution is chosen because it allows us to simplify the denominator using the trigonometric identity: 1 + tan²(θ) = sec²(θ).

3. Substitute and Simplify:

Now, substitute t and dt into the integral:

9 ∫ 1 / (81 + t²) dt = 9 ∫ 1 / (81 + (9 tan(θ))²) * (9 sec²(θ) dθ)

Simplify the expression:

9 ∫ (9 sec²(θ) dθ) / (81 + 81 tan²(θ)) = 9 ∫ (9 sec²(θ) dθ) / (81(1 + tan²(θ)))

Use the trigonometric identity 1 + tan²(θ) = sec²(θ):

9 ∫ (9 sec²(θ) dθ) / (81 sec²(θ))

Cancel out common factors:

9 ∫ (1/9) dθ = ∫ dθ

4. Integrate:

Now, integrate with respect to θ:

∫ dθ = θ + C

Where C is the constant of integration.

5. Back-Substitute:

We need to express the result in terms of the original variable t. From the substitution t = 9 tan(θ), we have:

tan(θ) = t/9

Therefore,

θ = arctan(t/9)

Substitute this back into the expression:

θ + C = arctan(t/9) + C

6. Evaluate the Definite Integral:

Now we have the antiderivative:

9 ∫ 1 / (81 + t²) dt = arctan(t/9) + C

To evaluate the definite integral ∫ 9 / (81 + t²) dt from 0 to ∞, we need to find the difference of the antiderivative at the upper and lower limits of integration:

[arctan(t/9)] from 0 to ∞ = lim (b→∞) arctan(b/9) - arctan(0/9)

As b approaches infinity, arctan(b/9) approaches π/2. arctan(0/9) is equal to 0.

Therefore,

lim (b→∞) arctan(b/9) - arctan(0/9) = π/2 - 0 = π/2

Scientific Explanation

The process we followed relies on fundamental principles of calculus and trigonometry. Here's a breakdown of the scientific justification:

• Integral as Area: The definite integral represents the area under a curve. Evaluating it gives us the numerical value of that area.
• Antiderivative and Fundamental Theorem of Calculus: Finding the antiderivative is the reverse process of differentiation. The Fundamental Theorem of Calculus states that the definite integral of a function can be evaluated by finding the difference of its antiderivative at the upper and lower limits of integration.
• Trigonometric Substitution: This technique is used to simplify integrals involving expressions of the form a² + x², a² - x², or x² - a². The goal is to use trigonometric identities to eliminate the square root or simplify the denominator. In our case, the substitution t = a tan(θ) allowed us to utilize the identity 1 + tan²(θ) = sec²(θ).
• Arctangent Function: The arctangent function, also known as the inverse tangent function, is the inverse of the tangent function. It gives the angle whose tangent is a given number. Its derivative is 1 / (1 + x²), which is closely related to the integrand we were dealing with. The integral of 1 / (a² + x²) is (1/a) * arctan(x/a), which is the basis for our solution.
• Limits and Infinity: Evaluating the definite integral with an infinite limit requires understanding the concept of limits. We replaced the infinity with a variable b, took the antiderivative, and then found the limit as b approached infinity. This allowed us to determine the value the function approaches as t becomes infinitely large. The arctangent function approaches π/2 as its argument approaches infinity.

Deeper Dive into Trigonometric Substitution

Trigonometric substitution is a powerful technique for simplifying integrals that involve expressions of the form a² + x², a² - x², or x² - a². The choice of substitution depends on the form of the expression:

• a² + x²: Use x = a tan(θ). This leads to a² + x² = a² sec²(θ).
• a² - x²: Use x = a sin(θ). This leads to a² - x² = a² cos²(θ).
• x² - a²: Use x = a sec(θ). This leads to x² - a² = a² tan²(θ).

The goal is to choose a substitution that allows you to simplify the expression under the square root or in the denominator, making the integral easier to evaluate. The substitution also introduces a factor of dx, which must be expressed in terms of dθ using the derivative of the substitution. After integrating with respect to θ, you must back-substitute to express the result in terms of the original variable x.

Common Mistakes and How to Avoid Them

Evaluating integrals, especially those involving trigonometric substitutions, can be tricky. Here are some common mistakes and how to avoid them:

• Forgetting the Constant of Integration: Always remember to add the constant of integration C when finding the indefinite integral. While it cancels out when evaluating definite integrals, it's crucial for understanding the general antiderivative.
• Incorrect Substitution: Choosing the wrong trigonometric substitution can make the integral more complicated. Make sure to select the substitution that matches the form of the expression in the integral.
• Forgetting to Change the Limits of Integration: When evaluating a definite integral using substitution, you can either back-substitute to express the antiderivative in terms of the original variable and then use the original limits of integration, or you can change the limits of integration to be in terms of the new variable. Forgetting to do either of these will lead to an incorrect answer.
• Incorrectly Evaluating Limits: When dealing with infinite limits, make sure you understand the behavior of the functions involved as the variable approaches infinity. Using L'Hôpital's rule or understanding the asymptotic behavior of functions can be helpful.
• Algebraic Errors: Simple algebraic errors can derail the entire solution. Double-check each step to ensure accuracy.
• Not Simplifying Before Integrating: Simplifying the integrand before attempting to integrate can often make the process much easier. Look for opportunities to factor, cancel, or use trigonometric identities.

Real-World Applications

While the integral ∫ 9 / (81 + t²) dt might seem abstract, integrals of similar forms arise in various real-world applications:

• Physics: Integrals are used extensively in physics to calculate quantities such as work, energy, and displacement. Integrals involving similar forms can appear in problems related to damped oscillations, electromagnetic fields, and quantum mechanics.
• Engineering: Engineers use integrals to design structures, analyze circuits, and model fluid flow. Integrals involving trigonometric functions and rational functions are common in these applications.
• Probability and Statistics: Integrals are used to calculate probabilities and expected values. Integrals similar to the one we evaluated can appear in the context of probability density functions and cumulative distribution functions.
• Finance: Integrals are used to calculate present and future values of investments, as well as to model financial risk.
• Computer Graphics: Integrals are used to calculate areas and volumes of complex shapes, which is essential for rendering realistic images.

Understanding how to evaluate integrals is a fundamental skill in many scientific and engineering disciplines.

Conclusion

Evaluating the integral ∫ 9 / (81 + t²) dt from 0 to ∞ requires a combination of techniques, including simplification, trigonometric substitution, integration, and evaluation of limits. By following the steps outlined in this article, you can successfully solve this type of integral. Understanding the underlying mathematical principles and avoiding common mistakes will help you master this skill and apply it to various real-world problems. The result of the evaluation, π/2, represents the area under the curve of the function 9 / (81 + t²) from 0 to infinity, a value that may have significant meaning depending on the context of the problem in which this integral arises. Mastering this technique is a valuable asset for anyone working in science, engineering, or mathematics.