Problem 82 - Thomas Calculus

Question

Evaluate each integral in Exercises \(71-82\) by using any technique you think is appropriate. $$ \int \frac{d x}{x \sqrt{3+x^{2}}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( -\ln \left| \frac{\sqrt{3}}{x} + \frac{x}{\sqrt{3+x^2}} \right| + C \).

Step 1: Identify the Substitution

To solve the integral \( \int \frac{dx}{x \sqrt{3+x^2}} \), we will use trigonometric substitution. We choose \( x = \sqrt{3} \tan(\theta) \). This leads to \( dx = \sqrt{3} \sec^2(\theta) \, d\theta \), and \( \sqrt{3+x^2} = \sqrt{3} \sec(\theta) \).

Step 2: Substitute and Simplify the Integral

Substituting \( x = \sqrt{3} \tan(\theta) \), the integral becomes: \[ \int \frac{\sqrt{3} \sec^2(\theta) \, d\theta}{\sqrt{3}\tan(\theta) \cdot \sqrt{3}\sec(\theta)} \] Simplifying it results in: \[ \int \frac{\sec(\theta)}{\tan(\theta)} \, d\theta \] which is equivalent to \( \int \frac{1}{\sin(\theta)} \, d\theta \).

Step 3: Change Integral to a Standard Form

Rewriting \( \frac{1}{\sin(\theta)} \), we recognize it as \( \csc(\theta) \). The integral becomes \( \int \csc(\theta) \, d\theta \), whose antiderivative is a standard form: \[ -\ln |\csc(\theta) + \cot(\theta)| + C \]

Step 4: Re-Substitute \( \theta \) Back to \( x \)

Since \( x = \sqrt{3} \tan(\theta) \), we have \( \tan(\theta) = \frac{x}{\sqrt{3}} \) and \( \sec(\theta) = \frac{\sqrt{3+x^2}}{\sqrt{3}} \). Therefore, \( \csc(\theta) = \frac{\sqrt{3}}{x} \) and \( \cot(\theta) = \frac{\sqrt{3}}{x} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{x \sqrt{3+x^2}} \). Rewrite the solution for \( \theta \) in terms of \( x \).

Step 5: Final Solution

Replacing \( \csc(\theta) \) and \( \cot(\theta) \) values, the integral becomes: \[ -\ln \left| \frac{\sqrt{3}}{x} + \frac{x}{\sqrt{3+x^2}} \right| + C \] This is the evaluated integral in terms of \( x \).

Key Concepts

Integration Techniques

Integration techniques are essential strategies for finding the integral of complex functions. In calculus, there are several methods to tackle different integral forms. One common method is trigonometric substitution, which is particularly useful for integrals involving square roots of expressions like \( a^2 + x^2 \), \( a^2 - x^2 \), and \( x^2 - a^2 \).

Trigonometric substitution uses trigonometric identities to transform a difficult integral into a more manageable one. For example, in the exercise provided, the substitution \( x = \sqrt{3} \tan(\theta) \) was used. This specific trigonometric substitution was chosen because it simplifies the square root term in the denominator. The transformation utilizes the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \), which aids in simplifying the integral.

Understanding when and how to use different integration techniques, such as trigonometric substitution, helps solve complex integrals and is a valuable skill in calculus. Remember:

• Identify the form of the integrand.
• Choose an appropriate substitution.
• Simplify the integral using identities.

This turns unknowns into manageable expressions and guides you to the integral’s antiderivative.

Definite and Indefinite Integrals

In calculus, integrals can be classified as either definite or indefinite, depending on whether they compute a fixed quantity or a general function. An indefinite integral gives a family of functions, known as antiderivatives, and is often represented with the integral sign followed by the differential (\( \int f(x) \, dx \)). This integral includes a constant of integration denoted by \( C \), because there’s an infinite number of antiderivatives differing by a constant.

Definite integrals, on the other hand, calculate the area under a curve in a given interval. They are represented as \( \int_a^b f(x) \, dx \) and result in a numerical value given by the difference of the antiderivatives at the boundaries \( a \) and \( b \).

While the exercise focuses on the indefinite integral \( \int \frac{dx}{x \sqrt{3+x^2}} \), understanding both definite and indefinite integrals is essential. The solution uses the indefinite form since it deals with antiderivatives and does not specify limits of integration. This means solving the integral results in a general form with \( C \). Recognizing when to use definite versus indefinite integrals can simplify problem solving and clarifications of results.

Antiderivatives

Antiderivatives are the reverse operation of differentiation, allowing us to determine a function whose derivative is the given function. They are fundamental to the concept of indefinite integrals. Solving an integral involves finding the antiderivative and adding a constant of integration \( C \). This process is well-demonstrated in the problem solution, where the trigonometric manipulation led to a known antiderivative form.

For instance, in the provided solution, the trigonometric substitution transformed the problem into \( \int \csc(\theta) \, d\theta \), which is a standard integral. Its antiderivative is \( -\ln |\csc(\theta) + \cot(\theta)| + C \).

The key steps for finding antiderivatives include:

• Identifying the function's standard form (e.g., pseudo-trigonometric identities).
• Looking up or knowing the antiderivative of similar forms.
• Applying the appropriate transformation backwards (like re-substituting \( \theta \) with in terms of \( x \)).

Understanding antiderivatives opens the door to evaluating integrals and applying them to real-world problems in physics, engineering, and beyond. Once you grasp the concept of antiderivatives, solving indefinite integrals becomes more intuitive and methodical.