Problem 27 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{0}^{1 / \sqrt{3}} \frac{t^{2}-1}{t^{4}-1} d t\)

Step-by-Step Solution

Answer

The integral evaluates to \(\pi/6\).

Step 1: Factor the Denominator

The denominator of the integrand is \(t^4 - 1\). This expression can be factored using the difference of squares method: \(t^4 - 1 = (t^2 - 1)(t^2 + 1)\). So, \(\frac{t^2 - 1}{t^4 - 1}\) can be rewritten as \(\frac{t^2 - 1}{(t^2 - 1)(t^2 + 1)}\).

Step 2: Simplify the Fraction

In the expression \(\frac{t^2 - 1}{(t^2 - 1)(t^2 + 1)}\), cancel the \(t^2 - 1\) terms to simplify it to \(\frac{1}{t^2 + 1}\).

Step 3: Integrate the Simplified Expression

Now that we have \(\int \frac{1}{t^2 + 1} \, dt\), recognize that \(\frac{1}{t^2 + 1}\) is the derivative of the arctangent function. Therefore, the antiderivative is \(\tan^{-1}(t) + C\).

Step 4: Evaluate the Definite Integral

Evaluate \(\left[\tan^{-1}(t)\right]_{0}^{1/\sqrt{3}}\). This means calculating \(\tan^{-1}(1/\sqrt{3}) - \tan^{-1}(0)\).

Step 5: Solve the Expression

\(\tan^{-1}(0) = 0\) because the tangent of 0 is 0. For \(\tan^{-1}(1/\sqrt{3})\), recognize that \(\tan(\pi/6) = 1/\sqrt{3}\), so \(\tan^{-1}(1/\sqrt{3}) = \pi/6\). Therefore, the integral evaluates to \(\pi/6\).

Key Concepts

Factorization

Factorization is a crucial step when working with integrals, especially with polynomial expressions. It involves breaking down a complex expression into simpler factors, making it easier to handle and solve. For example, in the expression \( t^4 - 1 \), we can apply the difference of squares. This technique works because any term of the form \( a^2 - b^2 \) can be factored into \( (a-b)(a+b) \).
This is helpful because it simplifies the original expression to \( (t^2 - 1)(t^2 + 1) \), providing a pathway to reduce the complexity of the integral.

• Step 1: Notice that \( t^4 - 1 \) mimics the difference of squares \( (t^2)^2 - 1^2 \).
• Step 2: Factor it into \( (t^2 - 1)(t^2 + 1) \), which further simplifies calculations.

By factorizing, you reduce expressions to their basic components, which is invaluable for both integration and algebraic manipulation.

Definite Integral

A definite integral calculates the area under a curve between certain bounds. It's written in the notation \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. In the case of our problem, these bounds are 0 and \( 1/\sqrt{3} \).
Definite integrals help us find exact numerical values rather than indefinite integrals that provide a general form with a constant of integration \( C \).

• Understanding the bounds: \( t = 0 \) to \( t = 1/\sqrt{3} \)
• Evaluate the simplified integral \( \int \frac{1}{t^2+1} \, dt \) over these bounds.

In summary, converting a function into a definite integral allows you to determine specific numerical results, which is useful in a range of mathematical contexts, including physics and engineering.

Arctangent Function

The arctangent function, denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \), is the inverse of the tangent function. It provides the angle whose tangent is a given number. This function is vital in the integration process.
In the problem, after simplification, we arrive at integrating \( \frac{1}{t^2 + 1} \), recognized as the derivative of the arctangent. Thus, its antiderivative is \( \tan^{-1}(t) + C \).

• Recognize \( \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) + C \)
• Use the arctangent to evaluate the integral between bounds.
• For example, \( \tan^{-1}(1/\sqrt{3}) \) results in \( \pi/6 \) because \( \tan(\pi/6) = 1/\sqrt{3} \).

Understanding the role of arctangent in integration enriches your mathematical toolkit, allowing you to effectively solve integrals involving inverse trigonometric functions.