Problem 18 - 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. \(\int \frac{d x}{\sqrt{x^{2}+4 x+5}}\)

Step-by-Step Solution

Answer

\(\tan^{-1}(x + 2) + C\)

Step 1: Identify the quadratic expression

First, we need to identify the quadratic expression within the square root, which is \(x^2 + 4x + 5\).

Step 2: Complete the square

Rewrite the quadratic expression \(x^2 + 4x + 5\) in the form \((x + a)^2 + b^2\).- Take half of the coefficient of \(x\), square it, and add/subtract it inside the expression: \(\left( \frac{4}{2} \right)^2 = 4\).- The expression becomes: \(x^2 + 4x + 4 + 1 = (x+2)^2 + 1\).

Step 3: Substitute the completed square into the integral

Now that we have \((x+2)^2 + 1\), substitute this expression back into the integral:\[ \int \frac{d x}{\sqrt{(x+2)^2 + 1}} \].

Step 4: Use trigonometric substitution

Set \(x + 2 = \tan(\theta)\), thus \(dx = \sec^2(\theta) d\theta\), and the integral becomes:\[ \int \frac{\sec^2(\theta) d\theta}{\sqrt{\tan^2(\theta) + 1}} = \int d\theta \] since \(\sqrt{\tan^2(\theta) + 1} = \sec(\theta)\).

Step 5: Evaluate the integral

The integral \(\int d\theta\) is straightforward and evaluates to \(\theta + C\).

Step 6: Back-substitute to original variable

Since \(x + 2 = \tan(\theta)\), we have \(\theta = \tan^{-1}(x + 2)\). Substitute back to get the solution in terms of \(x\):\[ \tan^{-1}(x + 2) + C \].

Key Concepts

Completing the Square

Completing the square is a technique used to simplify quadratic expressions and make them easier to work with. It involves rewriting a quadratic equation of the form \(ax^2 + bx + c\) into a perfect square trinomial plus or minus a constant.
1. **Identify "a" and "b"**: Focus on the terms with \(x\) — in our expression, "a" is 1 and "b" is 4.2. **Find the square**: Take the coefficient of \(x\) (which is 4), halve it to get 2, and square it resulting in 4.3. **Rewrite the expression**: Add and subtract this square inside the expression: \(x^2 + 4x + 4 + 1 = (x+2)^2 + 1\).
Completing the square transforms the quadratic into a more manageable expression. This is especially useful in calculus for integrating functions, as seen in this exercise.

Trigonometric Substitution

Trigonometric substitution is a powerful technique for evaluating integrals, particularly when dealing with expressions involving square roots. It's especially handy when working with expressions like \((x+a)^2\) or \(x^2\) under a square root.
- **When to use it**: Activate trigonometric substitution when you see derivatives of inverse trigonometric functions, like under square roots.- **Basic idea**: Replace variables with trigonometric identities. For instance, substitute \(x + a = \tan(\theta)\) when encountering \((x+a)^2 + 1\). This simplifies square roots to secant functions, \(\sec(\theta)\). - **Complete the transformation**: Substitute back into the integral with \(dx = \sec^2(\theta) d\theta\), converting integrals into more familiar trigonometric ones.
This method simplifies complex expressions, allowing the integral to be easier to evaluate. Ultimately, it helps in getting back to original variables to complete the solution process.

Indefinite Integrals

Indefinite integrals, central in calculus, represent a family of functions with an undefined constant, \(C\). Evaluating an indefinite integral allows the determination of the original function from its derivative.
- **Format**: Expressed as \(\int f(x) \, dx\), representing the antiderivative.- **Result interpretation**: The result of an indefinite integral is not a specific number but an expression that involves a constant, such as \(\theta + C\) in our trigonometric substitution instance.- **Substitution technique**: Integrals often require substitution, either variable or trigonometric, to simplify and solve.
In this exercise, the integral \(\int d\theta\) was evaluated to \(\theta + C\). The final step was substituting back to the original variable, leading to \(\tan^{-1}(x + 2) + C\). The indefinite integral thus represents a broad set of solutions, capturing all possible antiderivatives of the function.