Problem 47 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Substitution

First, look for a substitution that simplifies the integral. Notice that the expression under the square root is of the form \(3 + (\ln y)^2\). Let's set \(u = \ln y\).

Step 2: Express \(dy\) in terms of \(du\)

Once you have \(u = \ln y\), differentiate both sides with respect to \(y\): \(\frac{du}{dy} = \frac{1}{y}\). Thus, \(dy = y \cdot du\).

Step 3: Substitute in the Integral

Substitute \(u = \ln y\) and \(dy = y \cdot du\) into the integral. The integral then becomes \(\int \frac{y \cdot du}{y \cdot \sqrt{3+u^2}}\). The \(y\)'s cancel out, simplifying the integral to \(\int \frac{du}{\sqrt{3+u^2}}\).

Step 4: Identify the Table Integral

The integral \(\int \frac{du}{\sqrt{3+u^2}}\) is a standard one. Recognize it as an integral of the form \(\int \frac{1}{\sqrt{a^2+u^2}} du = \ln |u + \sqrt{u^2 + a^2}| + C\), where \(a = \sqrt{3}\).

Step 5: Evaluate the Integral Using the Table

Using the formula, the integral \(\int \frac{du}{\sqrt{3+u^2}}\) evaluates to \(\ln |u + \sqrt{3+u^2}| + C\). Substitute \(u = \ln y\) back into this expression.

Step 6: Substitute Back to \(y\)

Replace \(u\) with \(\ln y\) to get the answer in terms of \(y\): \(\ln |\ln y + \sqrt{3 + (\ln y)^2}| + C\).

Key Concepts

Substitution Method

In calculus, the substitution method is a handy tool used to make integration simpler. When you encounter a complex integral, this method can help simplify the expression by transforming it into a familiar form.

To apply substitution, the key is choosing a substitution that simplifies the integral. In our example, we noticed that the expression under the square root is \(3 + (\ln y)^2\). By setting \(u = \ln y\), the expression becomes more manageable.

• Differentiate to find \(dy\) in terms of \(du\): \(\frac{du}{dy} = \frac{1}{y}\).
• Rearrange to get \(dy = y \cdot du\).

This transformation makes it easier to evaluate the integral by substituting \(u\) and its differentials in place of the original variables. As a result, the integral turns into a simpler form that can be readily solved.

Integration Table

Integration tables are invaluable resources that provide solutions to standard integrals. They list integral forms alongside their derived solutions, making it quick to evaluate integrals when they match a listed form.

In the given example, after substitution, the integral \(\int \frac{du}{\sqrt{3+u^2}}\) matches a known entry in an integration table.

• A typical form listed is \(\int \frac{1}{\sqrt{a^2+u^2}} \, du = \ln |u + \sqrt{u^2 + a^2}| + C\)
• For our integral, \(a\) is \(\sqrt{3}\)

Once identified, simply use the corresponding solution from the table without delving into solving the integral manually. This step saves time and ensures accuracy.

Logarithmic Functions

Logarithmic functions are a type of mathematical function that is the inverse of exponential functions. The natural logarithm, denoted as \(\ln\), is particularly common in calculus.Using \(\ln\) in algebraic manipulations makes several expressions involving exponents simpler. In this exercise, setting \(u = \ln y\) transforms an initially complex problem into a standard form with manageable algebra.

• The function's properties facilitate simplifying expressions, differentiating, and integrating.
• In substitution, \(\ln\) redefines the variable to make calculations smoother.

Mastering the use of logarithmic functions aids in handling integrals, as seen from a simpler substitution leading to an accessible table integral format.