Question

Evaluate the given integral. $$ \int_{0}^{1 / 2} \frac{1}{1-t^{2}} d t $$

Short answer

The integral evaluates to \( \frac{1}{2} \ln(3) \).

Step-by-step solution

Recognize the Integral Form

The integral given, \( \int_{0}^{1/2} \frac{1}{1-t^2} \, dt \), is similar to the integral form of an inverse hyperbolic tangent function. The standard form to recognize is \( \int \frac{1}{1-x^2} \, dx = \text{artanh}(x) + C\).

Apply Formula and Evaluate

Using the recognition from Step 1, apply the formula. Evaluating the definite integral, substitute \( t = \frac{1}{2} \) and \( t = 0 \) into the antiderivative form:\[ \left[ \text{artanh}(t) \right]_{0}^{1/2} = \text{artanh}\left(\frac{1}{2}\right) - \text{artanh}(0) \]

Calculate Specific Values

Calculate \( \text{artanh}(0) \), which equals 0, because the hyperbolic arctangent of 0 is 0. Next, calculate \( \text{artanh}\left(\frac{1}{2}\right) \). Use properties or a calculator to find:\[ \text{artanh}\left(\frac{1}{2}\right) = \frac{1}{2} \ln \left( \frac{1 + 1/2}{1 - 1/2} \right) = \frac{1}{2} \ln(3) \]

Find the Result

Substitute the calculated values back:\[ \frac{1}{2} \ln(3) - 0 = \frac{1}{2} \ln(3) \]Thus, the definite integral evaluates to \( \frac{1}{2} \ln(3) \).

Key concepts

Inverse Hyperbolic Functions in Calculus

Inverse hyperbolic functions are useful tools in calculus, especially when dealing with integrals involving expressions like \( \frac{1}{1-x^2} \). These functions are analogous to the inverse trigonometric functions, but instead, they relate to hyperbolic functions. A common inverse hyperbolic function is the inverse hyperbolic tangent function, denoted as \( \text{artanh}(x) \). It is defined as the integration of the form \( \int \frac{1}{1-x^2} \, dx \). This corresponds directly to the standard hyperbolic tangent function, which forms a basis for handling specific integral problems like the one given in the exercise.
By recognizing the integral form as an inverse hyperbolic function, it simplifies the process significantly, meaning less manual computation and a more direct path to the solution.

Understanding Definite Integrals

Definite integrals are a fundamental concept in calculus. They calculate the net area under a curve, bounded by the given limits of integration. In our exercise, the definite integral is from 0 to \( \frac{1}{2} \). Unlike indefinite integrals, definite integrals provide a specific numerical value as a result.
When evaluating a definite integral, we first find the antiderivative of the integrand. Then, we use the Fundamental Theorem of Calculus to substitute the upper and lower limits into this antiderivative. This involves calculating the antiderivative at these two bounds and finding their difference.
This systematic approach not only gives us the net area but also reveals key properties of functions over specific intervals.

Integral Evaluation Steps Explained

Integral evaluation is often broken down into clear, logical steps to make complex problems more approachable. Here's a simplified outline of the steps used to evaluate the integral in the exercise:

• Recognize the Integral Form: Notice if the integral resembles any standard or known form, such as inverse hyperbolic functions. This recognition is crucial to applying known antiderivatives directly.
• Apply the Formula: Use the form recognized in step one to write the antiderivative. This step simplifies the evaluation by leveraging known results.
• Evaluate Definite Boundaries: Substitute the given limits into the antiderivative. Here, this involved calculating \( \text{artanh} \) at \( t = \frac{1}{2} \) and \( t = 0 \).
• Compute Results: Calculate numerical values if necessary and find the difference between them to determine the definite integral's value.

Each step ensures that the process remains structured, minimizing errors and providing a clean path to the final answer, \( \frac{1}{2} \ln(3) \) in this case.