Problem 78 - Applied Calculus

Question

Which two formulas can find \(\int \frac{1}{1-t^{2}} d t ?\)

Step-by-Step Solution

Answer

The integral can be expressed as either \( \text{artanh}(t) + C \) or \( \frac{1}{2} \ln\left|\frac{1+t}{1-t}\right| + C \).

Step 1: Identify the Integral Type

We are asked to find the integral \( \int \frac{1}{1-t^{2}} \, dt \). Notice that the integrand \( \frac{1}{1-t^2} \) resembles the derivative of the inverse hyperbolic tangent and the inverse trigonometric arctangent.

Step 2: Recognize the Standard Formulas

The integral of \( \frac{1}{1-t^2} \) can be solved using standard integral formulas. It is known that \( \int \frac{1}{1-u^2} \, du = \text{artanh}(u) + C \) and also that \( \int \frac{1}{1-x^2} \, dx = \text{arctan}\left(\frac{x}{\sqrt{1-x^2}}\right) + C \) for specific transformations.

Step 3: Apply Inverse Hyperbolic Function

For the hyperbolic function, use the relation \( \int \frac{1}{1-t^2} \, dt = \text{artanh}(t) + C \), where \( \text{artanh} \) is the inverse hyperbolic tangent.

Step 4: Consider a Different Approach (Partial Fraction for Trigonometric)

Alternatively, consider partial fraction decomposition: \( \frac{1}{1-t^2} = \frac{1/2}{1-t} + \frac{1/2}{1+t} \). Upon integrating these fractions, \( \int \frac{1}{1-t} \, dt = -\ln|1-t| + C \) and \( \int \frac{1}{1+t} \, dt = \ln|1+t| + C \). Combining results, \( \int \frac{1}{1-t^2} \, dt = \frac{1}{2} \ln\left|\frac{1+t}{1-t}\right| + C \), which relates to transforming it using \( \text{arctan}\).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of the familiar inverse trigonometric functions. They extend the concept of trigonometry to hyperbolic geometry. The inverse hyperbolic tangent, denoted as \( \text{artanh}(x) \), is one such function. It is the inverse of the hyperbolic tangent function, \( \text{tanh}(x) \). This means if \( y = \text{tanh}(x) \), then \( x = \text{artanh}(y) \). One key property of inverse hyperbolic functions is their definition over real numbers, unlike inverse trigonometric functions, which are often constrained to specific intervals.

The integral \( \int \frac{1}{1-t^2} \, dt \) can be directly solved using the inverse hyperbolic function relation:

• The formula is \( \int \frac{1}{1-u^2} \, du = \text{artanh}(u) + C \), where \( C \) is the constant of integration.
• This shows that the antiderivative of \( \frac{1}{1-t^2} \) is \( \text{artanh}(t) + C \).

Integral Calculus

Integral calculus is a core area of calculus that deals with integrals and their properties. An integral accumulates the total effect, such as area under a curve or total growth from a rate of change. The integral \( \int \frac{1}{1-t^2} \, dt \) examines the accumulative property of the function \( \frac{1}{1-t^2} \). Notably, in solving this, recognizing its connection to standard integrals greatly simplifies the process.

Key steps in solving integrals of this type involve:

• Identifying the form of the integrand, which is important because it determines which formula or technique applies.
• Applying known integral formulas or transformations to simplify or solve the integral directly.

In this specific example, the use of partial fraction decomposition can also be an insightful method. This involves breaking down \( \frac{1}{1-t^2} \) into simpler fractions that can be integrated more easily.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus for finding angles when given specific trigonometric values. They include functions like \( \text{arcsin}(x) \), \( \text{arccos}(x) \), and \( \text{arctan}(x) \). In the provided integral calculus problem, one alternative to solve \( \int \frac{1}{1-t^2} \, dt \) involves using the inverse tangent function, or \( \text{arctan}(x) \).

The formula corresponding to this approach is:

• \( \int \frac{1}{1-x^2} \, dx = \text{arctan}\left(\frac{x}{\sqrt{1-x^2}}\right) + C \).

When applying partial fraction decomposition, you end up with an expression that can similarly relate to the \( \text{arctan} \) function through transformations. This demonstrates the versatility of inverse trigonometric functions in integral calculus, offering multiple paths to obtain the integral's solution.