Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int \frac{d x}{\sqrt{16+6 x-x^{2}}}\)

Step-by-Step Solution

Answer

\(\int \frac{d x}{\sqrt{16+6x-x^{2}}} = \arcsin\left(\frac{x-3}{5}\right) + C\)

Step 1: Complete the Square

The expression under the square root is a quadratic expression: \(16 + 6x - x^2\). To make it easier to integrate, we will complete the square.\[-x^2 + 6x + 16 = -(x^2 - 6x - 16)\]First, rewrite \(x^2 - 6x\) in a form that allows us to complete the square. Take half of the coefficient of \(x\) (which is \(-6\)), square it, and add inside the parentheses. So, \((-6/2)^2 = 9\).\[x^2 - 6x = (x^2 - 6x + 9 - 9) = (x-3)^2 - 9\]Replace back into the original expression:\[-((x-3)^2 - 9 + 16) = -(x-3)^2 + 25\]So, \(16 + 6x - x^2\) can be rewritten as \(25 - (x-3)^2\). Reinsert into the integral:\[\int \frac{dx}{\sqrt{25 - (x-3)^2}}\]

Step 2: Use Trigonometric Substitution

Since the integral is now in the form \(\sqrt{a^2 - u^2}\), we can use a trigonometric substitution. Set \(x-3 = 5\sin(\theta)\), then \(dx = 5\cos(\theta)\,d\theta\).This choice uses \(a = 5\) with the identity \(a^2 - \sin^2(\theta) = \cos^2(\theta)\).Substitute and simplify:\[\int \frac{5\cos(\theta)\,d\theta}{\sqrt{25 - 25\sin^2(\theta)}} = \int \frac{5\cos(\theta)\,d\theta}{5\cos(\theta)} = \int d\theta\]

Step 3: Integrate

The integral \(\int d\theta\) is straightforward. It simply integrates to \(\theta + C\), where \(C\) is the constant of integration. \[\int d\theta = \theta + C\]

Step 4: Back-Substitute the Trigonometric Function

Substitute back the original variable using \(\sin(\theta) = \frac{x-3}{5}\). Find \(\theta\) in terms of \(x\):\[\theta = \arcsin\left(\frac{x-3}{5}\right)\]

Step 5: Write the Final Answer

Substitute \(\theta\) back into the result:\[\int \frac{d x}{\sqrt{16+6x-x^{2}}} = \arcsin\left(\frac{x-3}{5}\right) + C\]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify integrals involving square roots of quadratic expressions. It cleverly uses trigonometric identities to transform the integral into a form that is easier to evaluate. This works particularly well for integrals of the form \( \sqrt{a^2 - u^2} \), \( \sqrt{u^2 - a^2} \), and \( \sqrt{a^2 + u^2} \).

In our example, after completing the square the expression under the square root was rewritten as \( \sqrt{25 - (x-3)^2} \). This matches the structure \( \sqrt{a^2 - u^2} \), where \( a = 5 \) and \( u = x-3 \). To simplify, we set \( x-3 = 5\sin(\theta) \), leveraging the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

By doing this substitution:

• We set \( u = a \sin(\theta) \), so \( dx = 5 \cos(\theta) \, d\theta \).
• Our expression under the root becomes \( \cos^2(\theta) \), as \( 1 - \sin^2(\theta) = \cos^2(\theta) \).

After substitution, the integral greatly simplifies to \( \int d\theta \). This step is essential because it converts a potentially complex integral into a straightforward one.

Completing the Square

Completing the square is a method used to simplify quadratic expressions, making them easier to integrate or differentiate. The primary goal is to rewrite a quadratic expression in the form of \( (x-h)^2 + k \). This helps in identifying patterns and applying advanced mathematical techniques such as trigonometric substitution.

For the expression \( 16 + 6x - x^2 \), we first recognize it as a quadratic that we can rearrange:

• First, rearrange as \( -x^2 + 6x + 16 = -(x^2 - 6x - 16) \).
• To complete the square for \( x^2 - 6x \), take half of the coefficient of \( x \) (i.e., \(-6/2 = -3\)) and square it to get 9.
• Rewrite \( x^2 - 6x + 9 - 9 \) as \((x-3)^2 - 9 \).

By substituting back, the expression becomes \( 25 - (x-3)^2 \), a form that is simple to work with in subsequent steps for integration.

Definite Integrals

Definite integrals are used to calculate the exact area under a curve in a given interval and have both upper and lower limits. They differ from indefinite integrals, which do not have boundaries and include a constant of integration \( C \).

In the context of our problem, while we worked through an indefinite integral, understanding definite integrals is still important. When solving a definite integral, once the antiderivative is found, you substitute the upper and lower limits into the result, and compute the difference between these two values. This area calculation is key in applications like finding total change, areas of regions, and physical quantities like displacement and work where bounds are specified. Although our problem involved finding an indefinite integral, appreciating when and how definite integrals are used can enrich your understanding of calculus.