Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the method of completing the square, along with a trigonometric substitution if needed, to evaluate each integral. \(\int \sqrt{5-4 x-x^{2}} d x\)

Step-by-Step Solution

Answer

The evaluated integral is \[ \frac{9}{2}\left(\arcsin\left(\frac{x+2}{3}\right) + \frac{1}{2}\sin\left(2\arcsin\left(\frac{x+2}{3}\right)\right)\right) + C \].

Step 1: Rewrite the Quadratic Expression

The given integral is \( \int \sqrt{5 - 4x - x^2} \, dx \). Begin by rewriting the quadratic expression \(5 - 4x - x^2\). Reorder the terms to \(-x^2 - 4x + 5\) for completing the square.

Step 2: Complete the Square

To complete the square, factor out \(-1\) from the first two terms: \(-1(x^2 + 4x) + 5\). Add and subtract \(4\), the square of half the coefficient of \(x\), inside the bracket: \(-1((x+2)^2 - 4) + 5\) which simplifies to \(-((x+2)^2) + 9\). Thus, the expression becomes \(9 - (x+2)^2\).

Step 3: Substitution

Set \(x + 2 = 3 \sin \theta\) so that \((x+2)^2 = 9\sin^2\theta\) and thus \(dx = 3 \cos \theta \, d\theta\). The integral becomes \(\int \sqrt{9 - 9\sin^2\theta} \cdot 3\cos\theta \, d\theta\).

Step 4: Simplify the Integral

Use the trigonometric identity \(\cos^2\theta = 1 - \sin^2\theta\) to simplify \(\sqrt{9 - 9\sin^2\theta} = \sqrt{9\cos^2\theta} = 3\cos\theta\). The integral becomes \(\int 3\cos\theta \cdot 3\cos\theta \, d\theta = 9\int \cos^2\theta \, d\theta\).

Step 5: Solve the Integral

To integrate \(\int \cos^2\theta \, d\theta\), use the identity \(\cos^2\theta = \frac{1}{2}(1 + \cos(2\theta))\). This gives \(9\int \frac{1}{2}(1 + \cos(2\theta)) \, d\theta\) or \(\frac{9}{2} \left(\theta + \frac{1}{2}\sin(2\theta)\right) + C\).

Step 6: Back Substitution

Since \(x + 2 = 3 \sin \theta\), solve for \(\theta\): \(\sin \theta = \frac{x+2}{3}\). Thus, \(\theta = \arcsin \left(\frac{x+2}{3}\right)\). Substitute back to get the integral in terms of \(x\): \[ \frac{9}{2}\left(\arcsin\left(\frac{x+2}{3}\right) + \frac{1}{2}\sin\left(2\arcsin\left(\frac{x+2}{3}\right)\right)\right) + C \].

Key Concepts

Completing the Square

Completing the square is a widely used technique in calculus and algebra to simplify quadratic expressions, allowing us to solve equations or integrate functions. The goal is to rewrite a quadratic expression in the form

• \( ax^2 + bx + c \)

as a perfect square trinomial plus or minus a constant:

• \( (x + h)^2 + k \)

This transformation makes subsequent calculations, like integrations or finding roots, easier.

To complete the square:

• Identify the quadratic expression, reorder the terms if necessary, and factor out the leading coefficient if it's not 1.
• Take half of the coefficient of the linear term, square it, and add and subtract this value inside the expression.
• Factor the resulting perfect square trinomial.

For instance, with \(-x^2 - 4x + 5\), we reorder to \(-x^2 - 4x + 5\), then factor out \(-1\), adjust to fit the completing the square formula, and rewrite it as \(-( (x + 2)^2 ) + 9\). This process aids in preparing the expression for further integration steps.

Trigonometric Substitution

Trigonometric substitution is a powerful tool for evaluating integrals involving square roots of quadratic expressions. This method leverages trigonometric identities to simplify the integral into a form that is often much easier to evaluate.

The key concept involves substituting a trigonometric function for a variable, such as:

• \(x = a \sin \theta\)
• \(x = a \tan \theta\)
• \(x = a \sec \theta\)

Each substitution is suited for specific forms of expressions, primarily those containing \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\), respectively.

In our example, after completing the square, we are left with \(9 - (x+2)^2\). We can use the substitution \(x + 2 = 3 \sin \theta\) because it fits the form \(a^2 - x^2\). This substitution facilitates transforming the integrand into a form suitable for applying trigonometric identities, simplifying the overall integration process.

Integration Techniques

Integration techniques are essential for solving a wide range of definite and indefinite integrals. These methods enhance our ability to resolve complex integrals that do not succumb to basic rules of integration. When direct integration is tedious or impossible, one might consider:

• Substitution Method: Changing variables to simplify the integrand.
• Integration by Parts: Useful for integrals containing products of functions.
• Trigonometric Integrals: Integrals involving trigonometric functions.
• Trigonometric Substitution: Especially useful for integrals with square roots.

In the solved example, combining the substitution with completing the square and applying the identity \(\cos^2\theta + \sin^2\theta = 1\) allows us to simplify \(\sqrt{9 - 9\sin^2\theta}\) to \(3\cos\theta\). This simplification, using a fundamental identity, turns the integral into a manageable form:

• \(\int 3\cos\theta \cdot 3\cos\theta \, d\theta = 9\int \cos^2\theta \, d\theta\)

Transitioning through each step, each technique works together to streamline the integration process.

Quadratic Expressions

Quadratic expressions play a significant role in various fields of mathematics and science, frequently appearing in problems related to optimization, motion, and, importantly, calculus. These expressions typically take the form:

• \(ax^2 + bx + c\)

Understanding how to manipulate them is fundamental to solving a broad array of problems.

Key steps in working with quadratic expressions include:

• Reorder the terms if necessary, placing them in a standard form.
• Use techniques like completing the square to reveal the underlying structure of the expression.
• Factorize straightforward quadratics or approximate solutions when exact forms aren't feasible.

In a calculus context, recognizing these expressions enables the application of integrative techniques more effectively, as seen with the integral

• \(\int \sqrt{5 - 4x - x^2} \, dx\)

Rewriting the expression using completing the square, as demonstrated, turns it into a form that is approachable using trigonometric substitution, ultimately leading to the integration solution.