Problem 8 - Calculus Early Transcendentals

Question

Use a table of integrals to determine the following indefinite integrals. $$\int \frac{d x}{\sqrt{x^{2}-25}}$$

Step-by-Step Solution

Answer

Question: Determine the indefinite integral of $$\frac{1}{\sqrt{x^2 - 25}}$$. Answer: The indefinite integral of $$\frac{1}{\sqrt{x^2 - 25}}$$ is $$\int \frac{1}{\sqrt{x^2 - 25}}\,dx = \ln\left|x + \sqrt{x^2 - 25}\right| + C$$.

Step 1: Identify the integral formula

Compare the given integral with the standard formula $$\int \frac{1}{\sqrt{x^2 - a^2}}\,dx = \ln\left|x + \sqrt{x^2 - a^2}\right| + C$$. In our case, the integrand is $$\frac{1}{\sqrt{x^2 - 25}}$$, which means that $$a^2 = 25$$ and therefore $$a = 5$$.

Step 2: Apply the integral formula

Substitute $$a = 5$$ in the integral formula: $$\int \frac{1}{\sqrt{x^2 - 5^2}}\,dx = \ln\left|x + \sqrt{x^2 - 5^2}\right| + C$$.

Step 3: Simplify the result

Simplify the expression within the logarithm function: $$\int \frac{1}{\sqrt{x^2 - 25}}\,dx = \ln\left|x + \sqrt{x^2 - 25}\right| + C$$. Since this is an indefinite integral, our solution is complete, and we have: $$\int \frac{1}{\sqrt{x^2 - 25}}\,dx = \ln\left|x + \sqrt{x^2 - 25}\right| + C$$.

Key Concepts

Indefinite Integrals

Indefinite integrals, also referred to as antiderivatives, are a core concept in calculus. They represent a family of functions whose derivative is the integrand. Unlike definite integrals, which calculate the area under a curve within specific limits, indefinite integrals do not have limits of integration and thus include an arbitrary constant, often represented by 'C'.

This is because when we take the derivative of a constant, it results in zero. Therefore, any constant added to the function will disappear during differentiation. When solving indefinite integrals, it’s crucial to include this constant in your final answer. Omitting it can lead to misunderstandings since the solution can represent any of a spectrum of possible functions.

Understanding indefinite integrals lays the foundation for solving more complex integration problems, making it an essential skill in mathematics.

Integral Table

An integral table is a valuable tool in calculus, providing a list of standard integral forms and their solutions. It acts much like a dictionary for integrals, offering readily available solutions to frequently encountered problems in calculus.

An integral table can save significant time, especially as you face more complex functions that match standard forms in the table. When using an integral table, the first step is to identify the form your integrand (the function you want to integrate) most closely resembles. This might involve rewriting or rearranging the function.

• Compare the function with known formulas in the table.
• Identify any parameters, such as constants, you need to substitute into the formula.
• Apply the formula and solve the integral.

For example, in the given exercise, comparing the original function with a listed integral helped quickly arrive at the solution without laborious calculations.

Logarithmic Integration

Logarithmic integration refers to the process of integrating functions that result in a natural logarithm in their solution. This often occurs when you deal with integrals involving fractions where the denominator includes square roots or quadratic expressions.

In the case of the integral in our exercise, the form \( \int \frac{1}{\sqrt{x^2 - a^2}}\,dx \) leads to a solution involving the natural logarithm. Such integrals are straightforward once you recognize the form and apply the proper formula, as often found in integral tables.

To carry out logarithmic integration successfully:

• Recognize the structure of your function matches the form leading to a logarithm.
• Use the correct integral formula that includes the logarithmic expression.
• Ensure all constants are correctly integrated into the solution, which involves understanding the relationship: \( a^2 = 25 \) means \( a = 5 \) in the exercise.

Logarithmic integration highlights the beauty of calculus, turning seemingly complex functions into elegant solutions through proper technique.