Problem 79 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(71-94\) $$ \int_{0}^{2} \frac{d t}{8+2 t^{2}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{16} \).

Step 1: Simplify the Integral

First, we will simplify the function inside the integral. Notice that the denominator can be factored out to make it easier to integrate.The initial integral is:\[ \int_{0}^{2} \frac{d t}{8+2 t^{2}} \]We can factor out the common factor of 2 from the denominator:\[ \int_{0}^{2} \frac{d t}{2(4 + t^2)} \]So it simplifies to:\[ \frac{1}{2} \int_{0}^{2} \frac{d t}{4 + t^2} \].

Step 2: Recognize the Integral Formula

Now, recognize that the integral \( \int \frac{d t}{4 + t^2} \) is of the form \( \int \frac{d x}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C \), where \( a = 2 \).This gives us:\[ \int \frac{d t}{4 + t^2} = \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) + C \].

Step 3: Integrate and Apply Limits

Substitute this back into our integral and compute it over the limits from 0 to 2:\[ \frac{1}{2} \left[ \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \right]_{0}^{2} \].Simplifying, this becomes:\[ \frac{1}{4} \left[ \tan^{-1}\left(\frac{2}{2}\right) - \tan^{-1}(0) \right] \].This further simplifies to:\[ \frac{1}{4} \left[ \tan^{-1}(1) - \tan^{-1}(0) \right] \].

Step 4: Calculate the Numeric Result

Calculate the values from the trigonometric identities:- \( \tan^{-1}(1) = \frac{\pi}{4} \), because \( \tan(\frac{\pi}{4}) = 1 \).- \( \tan^{-1}(0) = 0 \), because \( \tan(0) = 0 \).Substitute these back with the simplified expression:\[ \frac{1}{4} \left( \frac{\pi}{4} - 0 \right) \].Thus, the answer is:\[ \frac{\pi}{16} \].

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus, focusing on finding the accumulated quantity, like areas under curves. Different techniques help solve a wide range of integrals, making it crucial to understand and identify the best method to use. Some of the common integration techniques include:

• Substitution Method: This involves replacing variables to simplify an integral, often by setting a part of the expression equal to a new variable.
• Integration by Parts: This technique is useful when integrals involve products of functions and follows from the product rule of differentiation.
• Partial Fractions: When dealing with rational functions, breaking them into simpler fractions helps in easier integration.
• Trigonometric Integrals: These involve trigonometric functions and sometimes require substitution with trigonometric identities.

In our exercise, we initially simplified the integrand by factoring out a common factor. This simplification set the stage for recognizing the use of a specific formula, helping to integrate the expression conveniently.

Definite Integrals

Definite integrals are concerned with finding the net area under a curve within particular limits or bounds. These integrals have upper and lower limits, specified in the integral notation as \( \int_{a}^{b} f(x) \, dx \).

• Properties: The fundamental theorem of calculus connects differentiation and integration by stating that if \( f \) is continuous on \([a, b] \), then \( \int_{a}^{b} f(x) \, dx \) equals the change in the antiderivative function \( F(x) \) over \([a, b] \).
• Calculation: After finding the indefinite integral (antiderivative), evaluate it at the upper and lower limits. Subtract these results to find the value of the definite integral.

In our solution, we computed the indefinite integral using a trigonometric identity and then applied the limits, 0 and 2, to finalize the value of the integral, showing the powerful relationship between the function and its accumulation over that specific interval.

Trigonometric Substitution

Trigonometric substitution is a technique used for integrating functions, especially when dealing with expressions involving square roots or where terms look similar to trigonometric identities.Common substitutions involve:

• \( x = a \sin(\theta) \) or \( x = a \cos(\theta) \) for \( a^2 - x^2 \), coming from the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
• \( x = a \tan(\theta) \) for \( a^2 + x^2 \), as in our solution, where this substitution simplifies the integral to a form solvable using \( \tan^{-1}(x) \).
• \( x = a \sec(\theta) \) for \( x^2 - a^2 \), aligning with the identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \).

In the provided exercise, recognizing the form \( a^2 + t^2 \) led to using the inverse tangent function, enabling us to solve the integral efficiently. Such substitutions transform difficult integrals into simpler ones, making the overall computation process more streamlined and manageable.