Problem 50 - Thomas Calculus

Question

Use a substitution to change the integral into one you can find in the table. Then evaluate the integral. \(\int \frac{3 d y}{\sqrt{1+9 y^{2}}}\)

Step-by-Step Solution

Answer

\(\ln |\sqrt{1 + 9y^2} + 3y| + C\)

Step 1: Identify the substitution

To simplify the integral \( \int \frac{3 \, d y}{\sqrt{1+9y^2}} \), notice that the expression \( 1 + 9y^2 \) suggests a trigonometric identity transformation. This is a typical form of \(1 + a^2x^2\), where a hyperbolic or trigonometric substitution may be effective. Let's substitute \( y = \frac{1}{3} \tan(\theta) \).

Step 2: Compute the differential

Given the substitution \( y = \frac{1}{3} \tan(\theta) \), find \( dy \) in terms of \( d\theta \). Differentiate both sides: \( dy = \frac{1}{3} \sec^2(\theta) \, d\theta \).

Step 3: Change the integral to new variable

Substitute \( y = \frac{1}{3} \tan(\theta) \) into the integral and replace \( dy \) with the expression found: \( dy = \frac{1}{3} \sec^2(\theta) \, d\theta \). Now compute:\[\int \frac{3}{\sqrt{1 + 9\left(\frac{1}{3}\tan(\theta)\right)^2}} \cdot \frac{1}{3} \sec^2(\theta) \, d\theta\]This simplifies to \( \int \sec(\theta) \, d\theta \).

Step 4: Integrate with respect to new variable

The integral \( \int \sec(\theta) \, d\theta \) can be found in a table of integrals, and it equals \( \ln |\sec(\theta) + \tan(\theta)| + C \), where \(C\) is the constant of integration.

Step 5: Re-substitute original variable

Recall from our substitution that \( \theta = \tan^{-1}(3y) \), leading to \( \tan(\theta) = 3y \) and \( \sec(\theta) = \sqrt{1 + 9y^2} \). Substitute back:\[\ln |\sqrt{1 + 9y^2} + 3y| + C\]This is the solution in terms of the original variable \(y\).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify certain integrals by replacing a variable with a trigonometric function. This method is particularly helpful when handling expressions that match the form of trigonometric identities, such as \(1 + a^2x^2\).
For the integral \(\int \frac{3}{\sqrt{1+9y^2}} \, dy\), recognizing the form \(1 + 9y^2\) suggests a trigonometric substitution like \(y = \frac{1}{3} \tan(\theta)\). This substitution leverages the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\).
The purpose of trigonometric substitution is to convert the integral into a more manageable form using trigonometric identities. By substituting \(y = \frac{1}{3} \tan(\theta)\), we can transform the integral into one involving \( heta\), simplifying the process.

Table of Integrals

The table of integrals is a valuable resource for quickly finding the integrals of common functions without having to perform integration from scratch each time. When using substitution methods, you often end up with standard forms that appear in these tables.
In the provided solution, after substituting and simplifying, the integral became \( \int \sec(\theta) \, d\theta \). This is a standard integral found in many integral tables:

• \( \int \sec(\theta) \, d\theta = \ln |\sec(\theta) + \tan(\theta)| + C \)

The table of integrals allows you to immediately write down this result, making the process much quicker and reducing the chance of error. It's essential to understand how to find and use these tables efficiently.

Integration Techniques

Integration techniques are methods developed to solve integrals that may not be immediately straightforward. They can range from basic techniques like substitution and integration by parts to more advanced methods like partial fraction decomposition.
In this exercise, two integration techniques come into play:

• **Substitution:** Initial substitution with \(y = \frac{1}{3} \tan(\theta)\) transforms the integral into trigonometric terms.
• **Utilizing Integral Tables:** Once the integral is in a simpler form, \( \int \sec(\theta) \, d\theta \), it can be evaluated using a table of integrals.

Learning and practicing various integration techniques is crucial in efficiently solving a wide array of integrals. Each technique has its own rules and situations where it is most effectively used, making them versatile tools in calculus.