Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \frac{d x}{\sqrt{16+6 x-x^{2}}} $$

Step-by-Step Solution

Answer

The integral evaluates to \(\arcsin \left(\frac{x-3}{5}\right) + C\).

Step 1: Complete the Square

First, complete the square for the expression inside the square root. The expression is \(16 + 6x - x^2\). We rewrite it as \(-x^2 + 6x + 16\) and factor out the negative sign to get \(-(x^2 - 6x) + 16\). Now, complete the square on \(x^2 - 6x\):\[x^2 - 6x + 9 - 9 = (x-3)^2 - 9\]So, the expression becomes:\[-((x-3)^2 - 9) + 16 = -(x-3)^2 + 25\] Thus, we have \(25 - (x-3)^2\). The integral becomes \[ \int \frac{dx}{\sqrt{25 - (x-3)^2}} \].

Step 2: Use Trigonometric Substitution

Perform a trigonometric substitution. Let \(u = x - 3\), then \(du = dx\). The expression will become \(\int \frac{du}{\sqrt{25 - u^2}}\), which resembles the identity \(a^2 - u^2 = a \cos^2 \theta\). Set \(u = 5\sin \theta\), then \(du = 5\cos \theta \, d\theta\), and the integral becomes:\[\int \frac{5\cos \theta \, d\theta}{\sqrt{25 - 25 \sin^2 \theta}} = \int \frac{5\cos \theta \, d\theta}{\sqrt{25 \cos^2 \theta}}\]Simplifying gives:\[\int d\theta = \theta + C\].

Step 3: Back Substitution

Revert the trigonometric substitution to return to the variable \(x\). From \(\sin \theta = \frac{u}{5}\), we have \(u = 5\sin \theta\) and therefore \(\theta = \arcsin \left(\frac{u}{5}\right)\). Recall that \(u = x - 3\), so:\[\theta = \arcsin \left(\frac{x-3}{5}\right)\]Thus, the solution to the integral is:\[\arcsin \left(\frac{x-3}{5}\right) + C\].

Key Concepts

Integration Techniques

Integration is a fundamental technique in calculus that involves finding the integral of a function. This process is essential in various applications like calculating areas, volumes, or solving differential equations. When faced with more complex integrals, like those involving square roots, special integration techniques can come in handy. In this specific problem, we're dealing with an integral of the form \(\int \frac{dx}{\sqrt{a^2 \pm x^2}}\).

Often these are handled through trigonometric substitution, which is a powerful tool to simplify the expression. But before applying this, we may need to simplify the function further via other techniques, such as completing the square. This simplification makes it possible to recognize patterns or identities, paving the way for substitution or other tactical maneuvers in integration.

Completing the Square

Completing the square is a technique used in algebra to transform a quadratic expression into a perfect square trinomial, which is useful for integrating expressions involving quadratic polynomials. In our integration problem, the expression under the square root is \(16 + 6x - x^2\).

The process starts by reorganizing the terms and factoring out the negative sign from the expression. This gives \(-x^2 + 6x + 16\), which becomes \(-(x^2 - 6x) + 16\).

• Next, focus on \(x^2 - 6x\), and find a number that makes this a complete square. This involves taking half of the linear term's coefficient (-6), squaring it, and adding/subtracting that value, which is 9 in this case.
• The expression \((x-3)^2 - 9\) allows us to re-write the original quadratic as \(-(x-3)^2 + 25\).

This transformed expression then permits the application of trigonometric substitutions since it now fits the form \(a^2 - (x-b)^2\). This simplification is crucial in transforming complex trigonometric integrals into more straightforward problems where standard formulas and substitutions can be applied.

U-Substitution

U-substitution is another vital method in calculus for simplifying the process of integration. It is particularly effective when an integral includes a function and its derivative. This technique aims to reduce an integral into a standard form by selecting an appropriate substitution. In this problem, even though the term u-substitution is mentioned, the idea aligns well with trigonometric substitution when preparing for the actual substitution phase.

As we simplify \( \int \frac{du}{\sqrt{25 - u^2}} \), we apply a trigonometric identity through the substitution \(u = 5\sin \theta\). This choice is inspired by seeing the resemblance to the identity \(a^2 - u^2\). Here is how it unfolds:

• \(du\) transforms to \(5\cos \theta \, d\theta\), simplifying our integral into a more workable form.
• This approach allows us to evaluate the integral as \(\int d\theta = \theta + C\).

Once integrated, we ensure to revert to the original variable by back-substituting, leading us to express the solution in terms of \(x\). The u-substitution sets the groundwork for conducting a more complex substitution, enabling easier integration of functions initially challenging to tackle head-on.