Problem 57 - Calculus Early Transcendentals

Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If \(x=4 \tan \theta,\) then \(\csc \theta=4 / x\) b. The integral \(\int_{1}^{2} \sqrt{1-x^{2}} d x\) does not have a finite real value. c. The integral \(\int_{1}^{2} \sqrt{x^{2}-1} d x\) does not have a finite real value. d. The integral \(\int \frac{d x}{x^{2}+4 x+9}\) cannot be evaluated using a trigonometric substitution.

Step-by-Step Solution

Answer

Question: Determine if the following statement is true or false: '\(\csc \theta = 4/x\) when \(x = 4\tan \theta\).' Answer: False Explanation: After solving for \(\csc \theta\), we found that \(\csc \theta = \frac{4}{x\cos\theta}\) which is not equal to \(\frac{4}{x}\). A counterexample is when \(\theta = 45^{\circ}\), \(x = 4\), \(\csc\theta = \sqrt{2}\), and \(4/x = 1\).

Step 1: Evaluate the expression:

First, rewrite the equation \(x = 4\tan \theta\) as \(\tan \theta = x/4\). Now, recall the trigonometric identity: \(\tan \theta = \sin \theta / \cos \theta\). Using this, \(x/4 = \sin \theta / \cos \theta\).

Step 2: Solve for \(\csc \theta\):

Recall that \(\csc \theta = \frac{1}{\sin \theta}\). To find \(\csc \theta\), we need to find the expression for \(\sin \theta\) from the given equation. Multiply both sides of the equation by \(\cos \theta\) to get \(\sin \theta = x\cos \theta / 4\). Hence, \(\csc \theta = \frac{1}{\sin\theta} = \frac{1}{x\cos\theta / 4}=\frac{4}{x\cos\theta} \neq \frac{4}{x}\). The statement is false.

Step 3: Counterexample:

Let \(\theta = \arctan(1) = 45^{\circ}\). Then, \(x = 4\tan\theta = 4\), and \(\csc\theta = \sqrt{2} \neq 4/x = 4/4 = 1\). b.

Step 4: Observe the function:

Notice that in the integrand, \(\sqrt{1-x^2}\) is the equation of a semicircle with radius 1. In the interval [1,2], the integrand takes imaginary values as the square root of a negative number. Therefore, the integral does not have a finite real value. The statement is true. c.

Step 5: Evaluate the integral:

Let's evaluate the integral \(\int_{1}^{2} \sqrt{x^{2}-1} dx\). This can be done using the trigonometric substitution \(x = \cosh u\), with \(dx = \sinh u du\). Upon substitution, the integral becomes \(\int_{\textrm{arccosh}(1)}^{\textrm{arccosh}(2)} \sqrt{\cosh^2 u - 1} \sinh u du\). Now, notice that \(\sqrt{\cosh^2 u - 1} = \sinh u\). Therefore, the integral becomes \(\int_{\textrm{arccosh}(1)}^{\textrm{arccosh}(2)} \sinh^2 u du\). This integral exists and has a finite real value. The statement is false. d.

Step 6: Evaluate the integral:

First, complete the square in the denominator: \(x^2+4x+9 = (x+2)^2+5\). Now, let's find a trigonometric substitution for the integrand. For such a substitution to exist, there needs to be a trigonometric identity of the form \(\tan^2u + 1=n\) for some constant \(n\). In this case, it would have to be \((x+2)^2+5 = (x+2)^2+1(1)\). Unfortunately, this is not possible. Therefore, the integral cannot be evaluated using a trigonometric substitution. The statement is true.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in mathematics, allowing us to express trigonometric functions in different forms. These identities are essential in simplifying expressions and solving equations. For instance, the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) helps connect tangent with sine and cosine.

Other key identities include:

• Pythagorean Identities: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Reciprocal Identities: \(\csc \theta = \frac{1}{\sin \theta}\), \(\sec \theta = \frac{1}{\cos \theta}\)
• Angle Sum and Difference: \(\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b\)

Using these identities wisely can help us navigate complex trigonometric problems, such as finding \(\csc \theta\) from \(x = 4 \tan \theta\). It's crucial to analyze the given expressions and apply the right identity to find the solution.

Definite Integrals

Definite integrals allow us to calculate the area under a curve within a specific interval. They are essential in calculus, providing insights into real-world phenomena like displacement and growth over time. The notation \( \int_{a}^{b} f(x) \, dx \) expresses the integral from \(a\) to \(b\).

When working with definite integrals:

• Ensure the integrand is defined: In statements involving \(\int_{1}^{2} \sqrt{1-x^{2}} \, dx\), we must check for imaginary values, since \(\sqrt{1-x^2}\) represents a semicircle.
• Watch for finite results: Ensure the area calculated is finite and real. Sometimes, due to the function's nature, integrals within certain intervals may not yield a finite result.

Knowing how to assess these integrals helps us confirm whether they have specific real values, like recognizing that some intervals may result in imaginary or undefined values.

Calculus Counterexamples

Counterexamples in calculus illustrate that certain statements may not universally hold true. These examples are crucial for understanding the limitations and scope of mathematical theorems.

In the context of integrals and substitutions:

• Show basic assumptions fail: For \(x = 4 \tan \theta\), showing that \(\csc \theta eq \frac{4}{x}\) using a specific angle \(\theta\) disproves a statement.
• Identify impossible substitutions: Evaluating \( \int \frac{dx}{x^2+4x+9} \) involves attempting trigonometric substitutions. Completing the square exposes why no suitable trigonometric identity fits.

These counterexamples not only clarify why a statement is false but also deepen our understanding of mathematical principles and their boundaries, highlighting the importance of careful analysis.