Problem 46 - Essential Calculus

Question

Evaluate the integral \(\int \frac{x^{2}}{x^{2}+9} d x\) (a) algebraically using \(x^{2}=\left(x^{2}+9\right)-9\) and (b) using trigonometric substitution Discuss the results.

Step-by-Step Solution

Answer

The integrals evaluated using both methods give the same result: \(x - 3 \arctan\left(\frac{x}{3}\right) + C\), where \(C\) is an arbitrary constant. Despite using different methods of substitution, the resulting antiderivative remains the same.

Step 1: Algebraic Substitution

Begin by substituting \(x^{2}=\left(x^{2}+9\right)-9\) into the integral. So, \(\int \frac{x^{2}}{x^{2}+9} dx = \int \frac{(\left(x^{2}+9\right)-9)}{x^{2}+9} dx \). This can be simplified to \(\int dx - \int \frac{9}{\left(x^{2}+9\right)} dx\).

Step 2: Solving the Integral Algebraically

The integral can then be solved to yield \(x - 3 \arctan\left(\frac{x}{3}\right) + C\), where \(C\) represents the constant of integration.

Step 3: Trigonometric Substitution

Next, use the trigonometric substitution approach, where \(x = 3 \tan \theta\), and \(dx = 3 \sec^2\theta d \theta\). Substituting these values yields the integral to be: \(\int \frac{(3\tan \theta)^2}{(3\tan \theta)^2 + 9} (3\sec^2\theta) d\theta\)

Step 4: Solving the Integral Trigonometrically

After simplifying the integral and solving, you obtain the same result as the algebraic method, \(x - 3 \arctan\left(\frac{x}{3}\right) + C' \). The constants of integration may have different values, but they are arbitrary constants.

Step 5: Analysis Of Results

As obtained in Steps 2 and 4, regardless of using algebraic substitution or trigonometric substitution, the resulting antiderivative is the same. Therefore, different methods applied to evaluate the integral can provide equivalent results. The choice of method used depends on preference or the way the exact integral simplifies with each method.

Key Concepts

Algebraic Substitution

Algebraic substitution in integral calculus involves replacing a part of the integrand with a simpler expression that makes the integration process more straightforward. In the context of our exercise with the integral \(\int \frac{x^{2}}{x^{2}+9} dx\), an astute observation is made: \(x^{2}\) can be expressed as \(x^{2}+9-9\). By substituting \(x^{2}\) in this way, the integral separates into two easier parts: \(\int dx\) and \( - \int \frac{9}{x^{2}+9} dx\).

The first part, \(\int dx\), simplifies to \(x\), as the integral of a constant rate of change refers to the variable itself. The second part requires a bit more work, but recognizing the denominator as a squared term plus a number indicates it may be related to the arctan function. This substitution is a clever algebraic maneuver that often turns a complex integral into a set of simpler ones, facilitating the overall solution. Substitutions like these can transform the terrain of integration, making a once steep climb into a gentle walk.

Trigonometric Substitution

Trigonometric substitution employs trigonometry to simplify complex integrals with square roots or squared terms plus or minus a constant. It's particularly useful when dealing with integrals containing \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\). In our exercise, trigonometric substitution involves introducing a variable \(\theta\) such that \(x = 3\tan\theta\).

The majestic thing about trigonometric identities is that they can transform the original integral into a form involving trigonometric functions that are much more approachable. An integral that involves \(\tan^2\theta\) and \(\sec^2\theta\) can be integrated using fundamental trigonometric integrals. This method takes advantage of the interrelationships between trigonometric functions to simplify the calculation. The process demands a familiarity with trigonometry, but, when mastered, it's like having a Swiss army knife for integrals—versatile and powerful.

Antiderivative

The antiderivative is at the heart of solving integrals. It's a function that, when differentiated, yields the function within the integral sign. Finding an antiderivative means identifying the original function whose rate of change is given by the function we are integrating. In simpler terms, if you're walking forwards by following the integrand, then taking an antiderivative is like walking backwards to where you started.

In the solution to our integral, after employing algebraic or trigonometric substitutions, we determined that \(x - 3\arctan\left(\frac{x}{3}\right)\) is the antiderivative of \(\frac{x^{2}}{x^{2}+9}\). This result is crucial, as it represents the function whose differentiation gives us back our original integrand. Understanding antiderivatives is key to unlocking the treasures of integral calculus—each one is akin to a secret password that opens doors within the mathematical landscape.

Constant of Integration

The constant of integration, often denoted by \(C\), is a fundamental part of the integration process, stemming from the fact that the derivative of a constant is zero. This means that when we find an antiderivative, there could be an infinite number of functions that differ only by a constant value which all share the same derivative. Therefore, when we integrate, we must add this constant of integration to account for all possible original functions.

In our example after performing the integration, we note \(C\) at the end of the expression. This \(C\) is the anchor allowing our antiderivative to capture a whole family of curves in the Cartesian plane, all of which are vertically shifted versions of one another. This subtle yet profound concept ensures that the canon of integrals is complete, embracing not just a single function but a multitude of possibilities unified by a simple constant—our integration constant \(C\).