Problem 42 - Thomas Calculus

Question

Evaluate each integral in Exercises \(37-42\) by completing the square and using a substitution to reduce it to standard form. $$ \int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sec^{-1}(x - 2) + C \).

Step 1: Complete the Square

The quadratic expression under the square root is \(x^2 - 4x + 3\). To complete the square, rearrange it in the form of \((x-a)^2 - b^2\). We start by adding and subtracting 4.\[x^2 - 4x + 3 = (x^2 - 4x + 4) - 1 = (x-2)^2 - 1\]

Step 2: Substitute the Quadratic Expression

To simplify the integral, use the substitution \( u = x - 2 \). Then \( du = dx \), and \( x = u + 2 \).Substitute these into the integral:\[\int \frac{1}{u \sqrt{(u+2)^2 - 4(u+2) + 3}} \, du\]Substituting from Step 1:\[\int \frac{1}{u \sqrt{u^2 - 1}} \, du\]

Step 3: Use Trigonometric Substitution

Let \( u = \sec(\theta) \), then \( du = \sec(\theta) \tan(\theta) \, d\theta \) and \( \sqrt{u^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta) \).The integral becomes:\[\int \frac{\sec(\theta) \tan(\theta) \, d\theta}{\sec(\theta) \tan(\theta)} = \int d\theta\]

Step 4: Integrate and Back-Substitute

Integrate with respect to \( \theta \):\[\theta + C\]Back-substitute for \( x \): From \( u = \sec(\theta) \) and \( u = x - 2 \), we have \( \sec(\theta) = x - 2 \) or \( \theta = \sec^{-1}(x - 2) \).Thus, the antiderivative is:\[\sec^{-1}(x - 2) + C\]

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to transform a quadratic expression into a perfect square trinomial. This simplifies expressions and helps solve problems involving quadratic equations.
By adding and subtracting the same value, we can rearrange a quadratic expression into the form \((x-a)^2 - b^2\).
Here’s how you can complete the square for a quadratic expression:

• Identify the quadratic expression. For example, consider \(x^2 - 4x + 3\).
• Focus on the \(x^2\) and \(-4x\) terms. To complete the square, find half the coefficient of the \(x\)-term, square it, and add that square inside the expression.
• In our example, half of \(-4\) is \(-2\); squaring it gives 4. So, add and subtract 4: \(x^2 - 4x + 3 = (x^2 - 4x + 4) - 1\).
• This gives \((x-2)^2-1\), a completed square form.

Once in this form, the quadratic is easier to handle, especially when integrating or solving equations.

Trigonometric Substitution

Trigonometric substitution is a powerful technique for evaluating integrals, especially when you encounter square roots of expressions like \(a^2 - x^2\), \(a^2 + x^2\), or \(x^2 - a^2\). This method replaces variables with trigonometric functions, simplifying the integration process.
Here is how you perform trigonometric substitution:

• For expressions like \(\sqrt{u^2 - 1}\), use the substitution \(u = \sec(\theta)\).
• Calculate \(du\) as \(\sec(\theta) \tan(\theta) \, d\theta\). Note that \(\sqrt{u^2 - 1} = \tan(\theta)\) because \(\sec^2(\theta) - 1 = \tan^2(\theta)\).
• Substitute these into the integral to simplify it. For example, \(\int \frac{1}{u \sqrt{u^2 - 1}} \, du\) becomes \(\int \frac{\sec(\theta) \tan(\theta) \, d\theta}{\sec(\theta) \tan(\theta)} = \int d\theta\). This is much easier to integrate.

After integration, replace \(\theta\) with the original variable to get your result back in terms of \(x\).

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial in calculus.
An **indefinite integral**, often shown as \(\int f(x) \, dx\), represents a family of functions \(F(x) + C\), where \(C\) is the constant of integration. This integral describes solutions to differential equations and usually doesn't have specific bounds.

On the other hand, a **definite integral**, denoted by \(\int_{a}^{b} f(x) \, dx\), computes the net area under the curve of \(f(x)\) from \(x = a\) to \(x = b\). It has specific lower and upper bounds, giving a numerical result instead of a function.

• If you want a comprehensive understanding of a process, usually start with the indefinite integral: \(\int f(x) \, dx\).
• For practical applications like calculating total area, use a definite integral, which eliminates the constant \(C\).

In solving problems, knowing whether to apply a definite or indefinite integral can also help you choose the correct technique or formula.