Problem 413 - Calculus Volume 1

Question

For the following exercises, find the antiderivatives for the functions. $$\int \frac{d x}{a^{2}-x^{2}}$$

Step-by-Step Solution

Answer

\( \frac{1}{2a} \ln \left| \frac{x+a}{x-a} \right| + C \)

Step 1: Identify the Formula

We notice that the integral, \( \int \frac{dx}{a^2 - x^2} \), can be solved using a standard formula. This formula states that \( \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{x+a}{x-a} \right| + C \), where \( a \) is a constant and \( C \) is the constant of integration.

Step 2: Confirm the Form

The given integral \( \int \frac{dx}{a^2 - x^2} \) matches the standard form since it is in the form of \( \frac{1}{a^2 - x^2} \cdot dx \). Here, \( a^2 - x^2 \) is a difference of squares which is exactly the form we need for the formula.

Step 3: Apply the Formula

We apply the formula. Substitute directly into the formula to get the antiderivative. Thus, we have: \[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{x+a}{x-a} \right| + C \].

Step 4: Write the Final Answer

Ensure to include the constant of integration \( C \) to account for all antiderivatives of the function. Therefore, the antiderivative is: \[ \frac{1}{2a} \ln \left| \frac{x+a}{x-a} \right| + C \].

Key Concepts

Integral Calculus

Integral calculus is a fundamental aspect of calculus that involves the process of finding antiderivatives or integrals of functions. An antiderivative of a function is essentially its original function before differentiation. Integral calculus helps us move in the opposite direction of differentiation, allowing us to find a function given its rate of change.

There are two main types of integrals: definite and indefinite integrals.

• Definite integrals calculate the area under a curve between two specific points.
• Indefinite integrals, such as the one in our exercise, provide a general form that represents a family of functions and include a constant of integration.

Using a standard formula in integral calculus, as highlighted in the exercise, speeds up the process of finding antiderivatives, especially when encountering familiar patterns in functions. By recognizing these patterns, solving integrals becomes more intuitive and less labor-intensive.

Standard Formulas

Standard formulas in integral calculus serve as invaluable shortcuts for solving integrals. By memorizing these, you can efficiently find antiderivatives without extensive calculation. Standard formulas tackle frequently encountered integrals, and our current exercise is a prime example. It uses the formula:\[ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{x+a}{x-a} \right| + C \]Knowing this formula allows us to solve the integral instantly.

These formulas capitalize on common patterns found in algebraic functions, such as the difference of squares in this case:

\[a^2 - x^2 = (a-x)(a+x) \]Recognizing such patterns enables us to efficiently apply the right standard formula for rapidly calculating an integral, making the process systematic and manageable.

Constant of Integration

The constant of integration, often represented by the symbol \( C \), is a vital component in the result of an indefinite integral. When you calculate an antiderivative, you're finding a family of functions, all of which differentiate to yield the same derivative.

Since differentiation of a constant is zero, antiderivatives can each have different constant values but still be valid, hence the addition of \( C \) in the integral's result.

For example, when you find the antiderivative of \( \int 1 \cdot dx = x + C \), the \( C \) accounts for any possible starting point along the y-axis.Ultimately, the constant of integration ensures completeness in representing the entire function family, acknowledging that without it, we would only describe a single function rather than the entire set. Always remember: forgetting the constant of integration in your answers means excluding infinitely many solutions.