Chapter 8: Problem 42
Evaluate each integral in Exercises \(37-42\) by completing the square and using a substitution to reduce it to standard form. $$ \int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}} $$
Short Answer
Expert verified
The integral evaluates to \( \sec^{-1}(x - 2) + C \).
Step by step solution
01
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.\[x^2 - 4x + 3 = (x^2 - 4x + 4) - 1 = (x-2)^2 - 1\]
02
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:\[\int \frac{1}{u \sqrt{u^2 - 1}} \, du\]
03
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:\[\int \frac{\sec(\theta) \tan(\theta) \, d\theta}{\sec(\theta) \tan(\theta)} = \int d\theta\]
04
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:\[\sec^{-1}(x - 2) + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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.
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.
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 \(C\).