Problem 54 - Calculus Early Transcendentals

Question

Use a computer algebra system to evaluate the following indefinite integrals. Assume that a is a positive real number. $$\int\left(y^{2}+a^{2}\right)^{-5 / 2} d y$$

Step-by-Step Solution

Answer

Question: Find the indefinite integral of the function $$\int\left(y^{2}+a^{2}\right)^{-5 / 2} d y$$ Solution: The indefinite integral of the function is $$-\frac{1}{3a^4(y^2+a^2)^{3/2}}+C$$.

Step 1: Rewrite the integrand

To get started, let's first rewrite the integrand in the following form: $$\frac{1}{\left(y^2 + a^2\right)^{5/2}}$$

Step 2: Perform u-substitution

We will perform u-substitution, which is often useful when dealing with integrals of this form. Let \(u = y^2 + a^2\). Therefore, \(du = 2ydy\). Now, we can rewrite the integral as: $$\frac{1}{2}\int\frac{1}{u^{5/2}}du$$

Step 3: Evaluate the integral using CAS

Now, we can use a computer algebra system (CAS) to solve the integral: $$\frac{1}{2}\int\frac{1}{u^{5/2}}du$$ The CAS will return the result: $$-\frac{1}{3a^4(y^2+a^2)^{3/2}}+C$$

Step 4: Substitute back to y

Finally, we substitute back to the original variable \(y\): $$-\frac{1}{3a^4(y^2+a^2)^{3/2}}+C$$ This is the result of the indefinite integral $$\int\left(y^{2}+a^{2}\right)^{-5 / 2} d y$$

Key Concepts

Computer Algebra System (CAS)

In the context of calculus, a Computer Algebra System (CAS) is an incredibly useful tool for students and mathematicians alike. It allows for the manipulation and evaluation of mathematical expressions using a variety of algorithms and methods. A CAS can carry out a range of operations such as solving equations, performing symbolic integrations and differentiations, simplifying expressions, and much more.

For indefinite integrals, a CAS can be particularly helpful as it can handle complex or laborious calculations that would be time-consuming or difficult to work out by hand. For example, a CAS could be used to integrate functions involving powers, trigonometric functions, exponentials, logarithms, and even to solve integrals symbolically when possible. Leveraging a CAS saves time, reduces computational errors, and allows students to focus on understanding the underlying concepts rather than getting bogged down in algebraic manipulation.

U-Substitution

U-substitution is a technique that simplifies the process of finding indefinite integrals, especially when the integrand is a composite function. This method involves choosing a new variable, commonly referred to as 'u', which represents a function of the original variable. By doing so, it transforms the integral into a form that is often easier to evaluate.

The choice of 'u' is crucial and typically involves identifying a portion of the integrand whose derivative also appears elsewhere in the integrand. In the case provided, setting u equal to y² + a² is strategic because its derivative, 2ydy, is present, which allows for a neat substitution and simplification. Remember that after integrating with respect to 'u', you need to substitute back the original variable to get the final answer in terms of 'y'.

Integral Evaluation

The process of integral evaluation involves finding the antiderivative or the indefinite integral of a function. In the given exercise, after applying the u-substitution method, we used a CAS to evaluate the integral. The evaluation stage is where we determine the value of the integral, often ending up with an antiderivative plus a constant of integration, represented as C.

Integral evaluation can sometimes be done by recognizing standard integral forms or by using integration techniques such as substitution, integration by parts, partial fraction decomposition, etc. The evaluation step confirms whether the technique applied has successfully transformed the integral into a solvable form and therefore is a crucial part of the integration process.

Integration Techniques

There are multiple techniques for evaluating integrals, known as integration techniques. In addition to u-substitution, other techniques include integration by parts, partial fractions, trigonometric integration, and trigonometric substitution. Each of these methods has its own set of rules and scenarios where it is most effective.

Integration by parts is useful particularly when the integrand is a product of two functions. The partial fractions technique is appropriate for rational functions, where the numerator's degree is less than the denominator's. For integrands involving trigonometric functions, trigonometric integration or substitution might be utilized. Knowing when and how to apply these techniques is vital for any calculus student and helps in efficiently solving a wide array of integral problems.