Problem 74 - Calculus Single Variable

Question

Calculate the given integral. $$ \int \frac{x}{\sqrt{16 x-x^{2}-48}} d x $$

Step-by-Step Solution

Answer

The integral is \(-\sqrt{16 - (x-8)^2} + 8 \cdot \arcsin \left(\frac{x-8}{4}\right) + C\).

Step 1: Complete the Square

The expression under the square root is a quadratic expression in the form of \(16x - x^2 - 48\). Reorganize and complete the square for this expression. Rewrite it as \(-x^2 + 16x - 48\), which is equivalent to \(-(x^2 - 16x) - 48\). Complete the square for \(x^2 - 16x\) by \(x^2 - 16x = (x - 8)^2 - 64\). Therefore, \(-((x - 8)^2 - 64) - 48 = - (x - 8)^2 + 64 - 48 = 16 - (x - 8)^2\). Thus, \(\sqrt{16x-x^2-48} = \sqrt{16-(x-8)^2}\).

Step 2: Substitution

Let \(u = x - 8\), which implies \(du = dx\) and \(x = u + 8\). Substitute these into the integral: \(\int \frac{x}{\sqrt{16 - (x-8)^2}} \, dx = \int \frac{u+8}{\sqrt{16-u^2}} \, du\).

Step 3: Simplify the Integral

Split the integral into two parts: \(\int \frac{u+8}{\sqrt{16-u^2}} \, du = \int \frac{u}{\sqrt{16-u^2}} \, du + \int \frac{8}{\sqrt{16-u^2}} \, du\). These can be solved separately.

Step 4: Solve the First Integral

The first integral \(\int \frac{u}{\sqrt{16-u^2}} \, du\) can be solved via the substitution \(v = 16 - u^2\), \(dv = -2u \, du\), or directly noticed as the derivative of the natural log. Thus the integral becomes half of \(v\): \(-\sqrt{16-u^2}\) or simplifies as \(-\sqrt{16-(x-8)^2}\).

Step 5: Solve the Second Integral

The second integral is \(\int \frac{8}{\sqrt{16-u^2}} \, du\) which is the form of the standard integral \(\int \frac{1}{\sqrt{a^2-u^2}} \, du = \arcsin \frac{u}{a}\) times the constant 8. Thus, it results in \(8 \cdot \arcsin \left(\frac{u}{4}\right)\). Since \(u = x - 8\), this becomes \(8 \cdot \arcsin \left(\frac{x-8}{4}\right)\).

Step 6: Combine the Results

The integral \(\int \frac{x}{\sqrt{16-(x-8)^2}} \, dx\) is the sum of the two parts calculated: \(-\sqrt{16-(x-8)^2} + 8 \cdot \arcsin \left(\frac{x-8}{4}\right) + C\), where \(C\) is the constant of integration.

Key Concepts

Completing the Square

When dealing with quadratic expressions, completing the square helps to simplify integration by transforming the expression into a perfect square. This method is particularly useful in integrals involving square roots. Consider the expression under the square root in our integral:

• We start with the quadratic form: \(16x - x^2 - 48\).
• Reorganize and factor out negative signs to make it easier to apply the square: \(-(x^2 - 16x) - 48\).
• To complete the square for \(x^2 - 16x\), rearrange it as \((x - 8)^2 - 64\).
• Substituting back, we simplify it to \(16 - (x - 8)^2\).

Completing the square aids in transforming the integral into a more familiar form, facilitating trigonometric substitution.

Trigonometric Substitution

In integrals involving square roots like \(\sqrt{16 - (x-8)^2}\), trigonometric substitution simplifies the expression using right triangle identities. Here’s how it works:

• Recognize the structure that resembles \(\sqrt{a^2 - u^2}\), which indicates using sine functions since \(\sin^2 \theta + \cos^2 \theta = 1\).
• Make the substitution \(u = x - 8\), aligning the expression with a trigonometric identity form.
• This leads to the integral form \(\int \frac{u+8}{\sqrt{16-u^2}} \, du\), representing a standard arcsine integral.
• Separate into two manageable integrals: one resembling the sine, another the arc-sine standard form.

Using trigonometric substitution transforms the integral into familiar forms that are easier to evaluate.

Integration Techniques

Different integration techniques are pivotal in solving complex integrals efficiently. Let's examine the techniques applied in this scenario:

• Begin by splitting the integral: \(\int \frac{u+8}{\sqrt{16-u^2}} \, du\) into \(\int \frac{u}{\sqrt{16-u^2}} \, du + \int \frac{8}{\sqrt{16-u^2}} \, du\).
• The first integral \(\int \frac{u}{\sqrt{16-u^2}} \, du\) can be recognized as the derivative of a logarithmic function, simplifying directly to \(\sqrt{16-u^2}\).
• The second integral \(\int \frac{8}{\sqrt{16-u^2}} \, du\) uses the arcsine formula \(\arcsin\left(\frac{u}{a}\right)\), specifically with \(a=4\).
• The results combine to form the solution: \(-\sqrt{16-(x-8)^2} + 8 \cdot \arcsin \left(\frac{x-8}{4}\right) + C\).

Understanding these techniques and when to apply them enables the solution of integrals that may initially appear daunting.