Problem 30 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{2}^{4} \frac{d t}{t \sqrt{t^{2}-4}} $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{1}{2} \).

Step 1: Identify the Integral Type

The given integral is \( \int_{2}^{4} \frac{d t}{t \sqrt{t^{2}-4}} \). This integral is of the form \( \int \frac{d t}{t \sqrt{a^2 - x^2}} \), where we need a trigonometric substitution for evaluation.

Step 2: Use Trigonometric Substitution

For integrals involving \( \sqrt{t^2 - a^2} \), we use the substitution \( t = a \sec(\theta) \). Here, \( a = 2 \), so we set \( t = 2 \sec(\theta) \). Consequently, \( dt = 2 \sec(\theta) \tan(\theta) \, d\theta \).

Step 3: Substitute and Simplify the Integral

Substitute \( t = 2 \sec(\theta) \) and \( dt = 2 \sec(\theta) \tan(\theta) \, d\theta \) into the integral. The integral becomes \( \int \frac{2 \sec(\theta) \tan(\theta) \, d\theta}{2 \sec(\theta) \sqrt{(2 \sec(\theta))^2 - 4}} \). The expression simplifies to \( \int \sin(\theta) \, d\theta \).

Step 4: Simplify the Expression Further

The term \( \sqrt{(2 \sec(\theta))^2 - 4} = \sqrt{4 \sec^2(\theta) - 4} = \sqrt{4(\sec^2(\theta) - 1)} = 2 \tan(\theta) \). The expression inside the integral is simplified to \( \int \sin(\theta) \, d\theta \) as \( \sin(\theta) = \frac{\tan(\theta)}{\sec(\theta)} \).

Step 5: Integrate with Respect to \( \theta \)

Integrate \( \int \sin(\theta) \, d\theta \) obtaining \( -\cos(\theta) + C \).

Step 6: Return to Original Variable

Since \( t = 2 \sec(\theta) \), then \( \cos(\theta) = \frac{2}{t} \). So, substitute back to get \( -\cos(\theta) = -\frac{2}{t} \). The antiderivative is \( -\frac{2}{t} + C \).

Step 7: Evaluate the Definite Integral

To evaluate \( \int_{2}^{4} -\frac{2}{t} \), compute \( \left[ -\frac{2}{t} \right]_{2}^{4} = -\frac{2}{4} + \frac{2}{2} = -\frac{1}{2} + 1 = \frac{1}{2} \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique often used to evaluate integrals involving square roots with quadratic expressions. For example, when we see an integral of the form \( \int \frac{1}{t \sqrt{t^2-a^2}} \,dt \), it's a hint to apply this method. In such cases, the substitution \( t = a \sec(\theta) \) can simplify the evaluation.

This transformation leverages the identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \), making square roots manageable.
By rewriting \( \sqrt{t^2 - 4} \) to \( \sqrt{4 \sec^2(\theta) - 4} \), it becomes \( 2 \tan(\theta) \).

Using this substitution, the difficulty of the square root is eliminated, allowing us to proceed further in the integration process.

Integral Calculation

Integral calculation involves transforming the original problem into a more workable form using substitution or other techniques. Here, after applying trigonometric substitution and converting to \( \theta \) terms, the integral simplifies significantly.

• We have \( dt = 2 \sec(\theta) \tan(\theta) \, d\theta \).
• And the integral \( \int \frac{2 \sec(\theta) \tan(\theta) \, d\theta}{2 \sec(\theta) \cdot 2 \tan(\theta)} = \int \sin(\theta) \, d\theta \).

This simplification is crucial as it leads us directly to an integral that can be easily evaluated without complex algebraic manipulations.

The new integral \( \int \sin(\theta) \, d\theta \) is one we've seen often, ready for immediate solution. This demonstrates the power of transforming integrals to match familiar forms.

Antiderivatives

In calculus, finding the antiderivative is the process of reversing differentiation. For the integral \( \int \sin(\theta) \, d\theta \), the antiderivative is straightforward.

The solution, \( -\cos(\theta) + C \), captures all functions whose derivative is \( \sin(\theta) \).

Once the antiderivative is found, we return from the substitution world back to the original variable by expressing \( \theta \) in terms of \( t \).
With \( \cos(\theta) = \frac{2}{t} \), we convert \( -\cos(\theta) \) to \( -\frac{2}{t} \).
This allows us to finalize the integral and evaluate the definite integral over the interval \([2, 4]\). Using the fundamental theorem of calculus, we substitute these limits giving us the exact area under the curve as \( \frac{1}{2} \).
Understanding this deep connection between substitution, antiderivatives, and their evaluation brings us a full circle in solving definite integrals.