Problem 34 - Calculus of a Single Variable

Question

Completing the Square In Exercises \(33-42,\) find or evaluate the integral by completing the square. $$ \int_{-2}^{2} \frac{d x}{x^{2}+4 x+13} $$

Step-by-Step Solution

Answer

The integral from -2 to 2 of \(\frac{d x}{x^{2}+4 x+13}\) is equal to \(\arctan{\frac{4}{3}}\)

Step 1: Complete the Square

We will start by completing the square in the denominator of the integral. Note that the square of \((x + 2)\) is \((x^2 + 4x + 4).\). Then the term 13 can be rewritten as \(4 + 9 = 13\). Thus, \(x^2 + 4x + 13\) can be rewritten as \((x + 2)^2 + 9\).

Step 2: Simplify the Integral

Replacing \((x^2 + 4x + 13)\) with \((x + 2)^2 + 9\), the integral \(\int_{-2}^{2} \frac{d x}{x^{2}+4 x+13}\) is rewritten as \(\int_{-2}^{2} \frac{d x}{(x+2)^{2}+ 9}\).

Step 3: Use Trigonometric Substitution

The integral \(\int_{-2}^{2} \frac{d x}{(x+2)^{2}+ 9}\) is a standard form for applying the trigonometric substitution \(x+2=3\tan{(\theta)}\), where \(\theta\) is a new variable. This way, \((x+2)^{2}+ 9= 3^2(\tan^2{(\theta)} + 1)=9\sec^{2}{(\theta)}\) and \(dx = 3\sec^{2}{(\theta)}d\theta\), simplifying the integral to be \(\int \frac{3\sec^{2}{(\theta)}d\theta}{9\sec^{2}{(\theta)}}\).

Step 4: Evaluate the Integral

The integral simplification in step 3 gives the integral \(\int d\theta\), which is equal to \(\theta\) up to a constant. But \(\theta\) in terms of x was \(\arctan{\frac{x+2}{3}}\). Then we need to evaluate it from -2 to 2.

Step 5: Evaluate at Limits

Evaluating \(\arctan{\frac{x+2}{3}}\) at the limits of -2 and 2 gives \(\arctan{\frac{2+2}{3}} - \arctan{\frac{-2+2}{3}} = \arctan{\frac{4}{3}} - \arctan{0}\). Since \(\arctan{0} = 0\), the integral from -2 to 2 of \(\frac{d x}{x^{2}+4 x+13}\) is equal to \(\arctan{\frac{4}{3}}\).

Key Concepts

Completing the Square

When you encounter a quadratic expression such as \(x^2 + 4x + 13\), completing the square is a useful technique that transforms the expression into a more manageable form. This method involves rewriting the expression as a perfect square trinomial plus a constant. In our example, \(x^2 + 4x + 13\) becomes \((x + 2)^2 + 9\).
This transformation is achieved by recognizing that the original quadratic in the denominator can be split into the square of a binomial and an additional constant:

• Start with the middle term, \(4x\), and divide the coefficient of \(x\) by 2, which gives 2.
• Square this result to obtain 4, then account for this arrangement by rewriting 13 as \(4 + 9\).
• Thus, the complete square expression becomes \((x + 2)^2 + 9\).

This makes the expression much easier for subsequent steps, particularly for integration.

Trigonometric Substitution

With the expression rewritten as \((x + 2)^2 + 9\), you can utilize trigonometric substitution to simplify the integration process. This particular form is conducive to a substitution involving a tangent function. Here, we set \(x+2=3\tan{\theta}\). This substitution leverages the trigonometric identity \(1 + \tan^2{\theta} = \sec^2{\theta}\), transforming the quadratic expression into a trigonometric one.
This substitution unfolds as:

• Calculate the derivative: \(dx = 3\sec^2{\theta}d\theta\).
• Replace the trinomial in the integral using \((x+2)^2 + 9 = 9\sec^2{\theta}\).
• Your integral then simplifies significantly, as you focus solely on basic trigonometric functions.

The power of trigonometric substitution lies in its ability to convert an otherwise complex algebraic expression into a simpler trigonometric one, easing the evaluation of the integral.

Integral Evaluation

Following trigonometric substitution, the integration simplifies to \(\int d\theta\). This represents one of the most straightforward forms of an integral, as the integral of \(d\theta\) is simply \(\theta\), plus a constant.
The challenge often lies in reverting back to the original variable, which involves:

• Recalling our substitution, \(x+2=3\tan{\theta}\), hence \(\theta = \arctan{(\frac{x+2}{3})}\).

This part of the process translates the solution back to the context of the original problem, providing a clear, evaluable function for the given limits of integration. Such clear and methodical integration simplifies the process and yields an understandable result.

Limits of Integration

Once the function has been returned to its original variable, it's important to evaluate it over the specified limits of integration, in this case from -2 to 2. With \(\theta = \arctan{(\frac{x+2}{3})}\), you substitute each limit into the solution:

• For the upper limit, substitute \(x = 2\), resulting in \(\theta = \arctan{(\frac{4}{3})}\).


• For the lower limit, substitute \(x = -2\), resulting in \(\theta = \arctan{(0)}\), which simplifies to 0.

The evaluated integral therefore becomes the difference: \(\arctan{(\frac{4}{3})} - \arctan{(0)}\). Considering \(\arctan{(0)}\) equals zero, the final result for the definite integral is \(\arctan{(\frac{4}{3})}\). Evaluating definite integrals serves to provide the exact area under the curve defined by the integral's function, within the specified bounds.