Problem 24 - 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. \(\int \frac{x}{\sqrt{4 x-x^{2}}} d x\)

Step-by-Step Solution

Answer

\(2\sin^{-1}\left(\frac{x-2}{2}\right) - (x - 2) + C\)

Step 1: Express the Integral in a Simplified Form

Given the integral \( \int \frac{x}{\sqrt{4x - x^2}} \, dx \), the first step is rewriting the expression inside the square root. Notice that \(4x - x^2\) can be rewritten as a complete square. Factoring out \(-1\), we have \(- (x^2 - 4x)\).

Step 2: Complete the Square Inside the Square Root

To complete the square for the expression \(x^2 - 4x\), take half of the coefficient of \(x\), which is -4, divide by 2 to get -2, and then square it to get 4. So, \(x^2 - 4x\) becomes \((x - 2)^2 - 4\). Therefore, \(4x - x^2\) can be rewritten as \(4 - (x - 2)^2\).

Step 3: Set Up the Trigonometric Substitution

Use the substitution \(x - 2 = 2\sin(\theta)\). This implies \(dx = 2\cos(\theta) d\theta\). The integral \(\int \frac{x}{\sqrt{4x - x^2}} \, dx\) is transformed. From the substitution, \(x = 2 + 2\sin(\theta)\), and \(\sqrt{4 - (x - 2)^2} = \sqrt{4 - 4\sin^2(\theta)} = 2\cos(\theta)\).

Step 4: Substitute Back to Transform the Integral

Substituting these expressions back into the integral gives \(\int \frac{2 + 2\sin(\theta)}{2\cos(\theta)} \cdot 2\cos(\theta) \, d\theta\). Simplifying, the integral becomes \(\int (2 + 2\sin(\theta)) \, d\theta\).

Step 5: Integrate with Respect to Theta

Now, integrate each term separately: \(\int 2 \, d\theta + \int 2\sin(\theta) \, d\theta\). The integral of the constant 2 is \(2\theta\), and the integral of \(2\sin(\theta)\) is \(-2\cos(\theta)\). Thus, the integral becomes \(2\theta - 2\cos(\theta) + C\).

Step 6: Substitute Back to Original Variable

Recall \(\theta\) was defined such that \(x - 2 = 2\sin(\theta)\), which implies \(\sin(\theta) = \frac{x-2}{2}\). To express \(\theta\) in terms of \(x\), use \(\theta = \sin^{-1}\left(\frac{x-2}{2}\right)\). For \(\cos(\theta)\), use the Pythagorean identity: \(\cos(\theta) = \sqrt{1-\sin^2(\theta)} = \sqrt{1-\left(\frac{x-2}{2}\right)^2}\). Simplifying gives \(\cos(\theta) = \frac{\sqrt{4 - (x-2)^2}}{2}\). Substitute \(\theta\) and \(\cos(\theta)\) back: \(2\sin^{-1}\left(\frac{x-2}{2}\right) - (x-2) + C\).

Step 7: Final Expression of the Integral

The integral evaluated is \(2\sin^{-1}\left(\frac{x-2}{2}\right) - (x - 2) + C\).

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to transform a quadratic expression into a perfect square plus or minus a constant. This process is particularly useful when you need to evaluate integrals involving expressions like \(x^2 - 4x\). To complete the square for this expression:

• Take the coefficient of \(x\), which is -4, divide it by 2 to get -2.
• Square the result to obtain 4.

This transforms \(x^2 - 4x\) into \((x - 2)^2 - 4\). This manipulation simplifies the expression under the square root, allowing us to address it more easily in integral calculus problems. Once the square is completed, analyzing and transforming the integral becomes more straightforward.

Trigonometric Substitution

Trigonometric substitution is a technique that involves replacing variables with trigonometric functions to simplify integrals, especially those involving square roots. In our exercise, after completing the square, the expression \(4x - x^2\) is rewritten as \(4 - (x-2)^2\). This expression closely resembles the Pythagorean identity \(1 - \sin^2(\theta) = \cos^2(\theta)\).

The substitution \(x - 2 = 2\sin(\theta)\) is used. This leads to \(dx = 2\cos(\theta) \, d\theta\). Simplifying further, we find \(\sqrt{4 - (x-2)^2} = 2\cos(\theta)\). By substituting these back into the integral, the complexity of dealing with a square root is significantly reduced, transforming it into a more manageable trigonometric integral.

Definite Integral

A definite integral is an integral with specified upper and lower limits. It represents the net area under the curve between those two points. In exercises dealing with definite integrals, the result is a specific numerical value rather than an algebraic expression with an arbitrary constant.

Evaluating a definite integral involves substituting the bounds after performing integration. For example, if evaluating from \(a\) to \(b\), after finding the antiderivative \(F(x)\), you compute \(F(b) - F(a)\). This process highlights the main difference between definite and indefinite integrals, with the former being concerned with finding an exact numerical result.

Indefinite Integral

Indefinite integrals are integrations without specified limits, resulting in an expression that includes an arbitrary constant \(C\). This symbol \(C\) represents the family of all antiderivatives of the function, providing an expression relating the integral of a function to its derivative.

• For example, the indefinite integral of \(f(x) = 2\) is \(F(x) = 2x + C\), representing all possible antiderivatives.
• Similarly, if \(f(x) = \sin(x)\), its indefinite integral is \(-\cos(x) + C\).

The arbitrary constant \(C\) reflects the infinite vertical shifts possible for the antiderivative graph, emphasizing the indefinite nature of these integrals. In our context, understanding indefinite integrals allows for manipulating and simplifying expressions post-integral transformation.