Problem 71 - Precalculus: Functions and Graphs

Question

Make the trigonometric substitution $$x=a \tan \theta \quad \text { for }-\pi / 2<\theta<\pi / 2 \text { and } a>0.$$ Simplify the resulting expression. $$\frac{1}{x^{2}+a^{2}}$$

Step-by-Step Solution

Answer

\(\frac{\cos^2 \theta}{a^2}\)

Step 1: Understanding the Substitution

We are asked to use the substitution \(x = a \tan \theta\) where \(-\pi/2 < \theta < \pi/2\) and \(a > 0\). This substitution is often used to simplify expressions involving \(x^2 + a^2\) because of the trigonometric identity \(\tan^2 \theta + 1 = \sec^2 \theta\).

Step 2: Substitute the Values

Substituting \(x = a \tan \theta\) into the expression we get \(x^2 = (a \tan \theta)^2 = a^2 \tan^2 \theta\). Therefore, \(x^2 + a^2 = a^2 \tan^2 \theta + a^2\).

Step 3: Simplify the Expression

Using the trigonometric identity \(\tan^2 \theta + 1 = \sec^2 \theta\), we can rewrite \(a^2 \tan^2 \theta + a^2\) as \(a^2(\tan^2 \theta + 1) = a^2 \sec^2 \theta\). Thus, the expression simplifies to \(\frac{1}{a^2 \sec^2 \theta}\).

Step 4: Further Simplify Using Trigonometric Identities

Since \(\sec \theta = \frac{1}{\cos \theta}\), we have \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\). Therefore, we can write \(\frac{1}{a^2 \sec^2 \theta} = \frac{1}{a^2} \cos^2 \theta\).

Step 5: Final Expression

The original expression \(\frac{1}{x^2 + a^2}\) simplifies to \(\frac{\cos^2 \theta}{a^2}\) using the substitution \(x = a \tan \theta\).

Key Concepts

Trigonometric Identities

When working with expressions that involve trigonometric functions, understanding trigonometric identities is vital. These identities are equations involving trigonometric functions that are true for any value of the variable.

• A common identity used in trigonometric substitution is: \( an^2 \theta + 1 = \sec^2 \theta\). This identity is immensely helpful when simplifying expressions that include a sum of squares.
• Another key identity to remember is: \( \sec \theta = \frac{1}{\cos \theta} \), which shows the relationship between secant and cosine functions. By recognizing and applying these identities, you can transform complicated expressions into simpler forms.

Using these identities allows you to simplify complex expressions by substituting trigonometric functions. In the context of the exercise, using \(x = a \tan \theta\) helps streamline the expression into a more manageable form.

Simplifying Expressions

Simplifying expressions involves rewriting them in their simplest form. This process can involve using identities, cancellations, or algebraic manipulations. Here, we harness trigonometric identities to simplify an expression that initially looks quite complex. Starting with the substitution \(x = a \tan \theta\), we replace \(x^2 + a^2\) with \(a^2 \tan^2 \theta + a^2\). Then, leveraging the identity \( \tan^2 \theta + 1 = \sec^2 \theta\), we transform it into \( a^2 \sec^2 \theta \). By further recognizing that \( \sec^2 \theta = \frac{1}{\cos^2 \theta}\), we bring the expression to \( \frac{1}{a^2 \sec^2 \theta} = \frac{1}{a^2} \cos^2 \theta\).Ultimately, our complicated expression simplifies down significantly to \( \frac{\cos^2 \theta}{a^2} \), showing the power of technique and identity usage in simplifying expressions.

Precalculus

Precalculus serves as a bridge between algebra and calculus, providing the foundational skills needed for more advanced mathematics. It encompasses various concepts, including trigonometry, which is essential for understanding calculus-related problems.

• In precalculus, trigonometry helps us understand angles, triangles, and the behavior of periodic functions.
• Trigonometric substitution, like \(x = a \tan \theta\), falls into this category as a technique used to simplify integrals and solve complex equations.
• This substitution technique leverages trigonometric identities to reframe algebraic expressions, making them easier to work with. Understanding these concepts in precalculus sets the stage for solving calculus problems that involve integration and limits.

Through mastering precalculus, including its trigonometric components, students become equipped to tackle a range of mathematical challenges, paving the way for the study of calculus.