Problem 25 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{9 d u}{1+9 u^{2}} $$

Step-by-Step Solution

Answer

The integral evaluates to \(3\tan^{-1}(3u) + C\).

Step 1: Identify Substitution

First, we need to identify a suitable substitution to simplify the integral. Notice the expression inside the integral, \(1 + 9u^2\), resembles a form related to the tangent trigonometric identity \(1 + \tan^2(x) = \sec^2(x)\). We can use the substitution \(u = \frac{1}{3}\tan(\theta)\).

Step 2: Express du in Terms of dθ

The substitution \(u = \frac{1}{3}\tan(\theta)\) implies \(du = \frac{1}{3}\sec^2(\theta)d\theta\). Now replace \(du\) in the integral with \(\frac{1}{3}\sec^2(\theta)d\theta\).

Step 3: Substitute and Rewrite the Integral

Substitute the expressions for \(u\) and \(du\) in the integral: \(\int \frac{9}{1 + 9u^2}du = \int \frac{9}{1 + 9 \left(\frac{1}{3}\tan(\theta)\right)^2} \cdot \frac{1}{3}\sec^2(\theta)d\theta.\)This simplifies to \(\int \frac{9}{1 + \tan^2(\theta)} \cdot \frac{1}{3}\sec^2(\theta)d\theta\).

Step 4: Simplify the Integral Using Trigonometric Identities

Using the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\), substitute in the integral: \(\int \frac{9}{\sec^2(\theta)} \cdot \frac{1}{3}\sec^2(\theta)d\theta = \int 3d\theta.\)

Step 5: Integrate with Respect to θ

The integral \(\int 3d\theta\) is straightforward to compute, resulting in \(3\theta + C\), where \(C\) is the constant of integration.

Step 6: Convert Back to Original Variable u

From the substitution, \(u = \frac{1}{3}\tan(\theta)\), thus \(\theta = \tan^{-1}(3u)\). Substitute back to express the final answer in terms of \(u\): \(3\theta = 3\tan^{-1}(3u)\).

Step 7: Final Step: Write the Solution

The evaluated integral is \(3\tan^{-1}(3u) + C\).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique often used in calculus to simplify the integration of expressions containing squares and square roots. This method involves replacing expressions with trigonometric functions to take advantage of their identities.

For example, when you see integrals like \( \int \frac{9 du}{1+9 u^2} \), this suggests the use of the tangent substitution because it closely resembles the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \).

The substitution we've used is \( u = \frac{1}{3}\tan(\theta) \). This not only simplifies the integral but also transforms it into a format that's easier to integrate further.

• This specific substitution allows the integral to be rewritten using trigonometric identities.
• It helps in recognizing patterns and replacing them appropriately.

After substitution and using trigonometric identities, many complex integrals become more straightforward to solve.

Calculus Problem Solving

Solving calculus problems with substitution methods, like trigonometric substitution, requires identifying opportunities to simplify expressions. Initially, always look for a pattern resembling a trigonometric identity.

In our example, we started by noticing the \( 1 + 9u^2 \) pattern, suggesting a trigonometric identity is usable. Rewriting the integral \( \int \frac{9}{1 + 9u^2} du \) using \( u = \frac{1}{3}\tan(\theta) \) is a critical first step in transforming the calculus problem into something more manageable.

• Break the problem into smaller steps.
• Apply appropriate substitutions that convert the integral into a recognized form.

Once substitutions are made, simplifying using trigonometric identities is crucial for reaching the solution. Problem-solving in calculus is about finding paths to simplify until we reach straightforward computations.

Definite and Indefinite Integrals

Integrals, in calculus, are categorized into definite and indefinite types. Understanding them is essential for solving a wide array of problems.

**Indefinite Integrals:** These involve integrating expressions without specific limits, producing a family of functions including a constant of integration \( C \). For instance, \( \int 3 d\theta = 3\theta + C \) represents an indefinite integral. The solution is not confined to a specific range, hence the constant \( C \) allows for any vertical shift of the function.

**Definite Integrals:** In contrast, definite integrals calculate the net area under a curve between specific bounds. They do not include a constant \( C \). However, this exercise primarily involves indefinite integrals, thus focusing more on understanding the function's behavior over its entire domain.

Learning to handle both types of integrals strengthens problem-solving skills in calculus, as each type signifies different real-world applications and solutions.