Problem 10 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-28\). $$ \int \frac{5 d x}{\sqrt{25 x^{2}-9}}, \quad x>\frac{3}{5} $$

Step-by-Step Solution

Answer

\( \ln \left| 5x + \sqrt{25x^2 - 9} \right| + C' \)

Step 1: Identify the Integral Form

We are given the integral \( \int \frac{5 \, dx}{\sqrt{25x^2 - 9}} \). This integral resembles the standard form \( \int \frac{dx}{\sqrt{a^2 x^2 - b^2}} \). We need to adjust our given integral to match this form as much as possible.

Step 2: Simplify the Denominator

Recognize that \( 25x^2 = (5x)^2 \) and \( 9 = 3^2 \). Thus, the expression under the square root is \( \sqrt{(5x)^2 - 3^2} \). This fits the integral form \( \int \frac{dx}{\sqrt{a^2 x^2 - b^2}} \) with \( a = 5 \) and \( b = 3 \).

Step 3: Perform a Trigonometric Substitution

To simplify this integral, use the substitution \( x = \frac{3}{5} \sec \theta \), which leads to \( dx = \frac{3}{5} \sec \theta \tan \theta \, d\theta \). Substitute \( x \) in the integral to transform it into a trigonometric form.

Step 4: Substitute and Simplify

Substitute \( x = \frac{3}{5} \sec \theta \) into the integral: \[ \int \frac{5 \, \cdot \left(\frac{3}{5} \sec \theta \tan \theta \right) \, d\theta}{\sqrt{(5 \cdot \frac{3}{5} \sec \theta)^2 - 9}} \] This simplifies to: \[ \int \frac{3 \sec \theta \tan \theta \, d\theta}{\sqrt{9 \sec^2 \theta - 9}} \] Simplifying further, we get: \[ \int \frac{3 \sec \theta \tan \theta \, d\theta}{\sqrt{9(\sec^2 \theta - 1)}} = \int \frac{3 \sec \theta \tan \theta \, d\theta}{3 \tan \theta} \] Which simplifies to: \[ \int \sec \theta \, d\theta \]

Step 5: Integrate

The integral \( \int \sec \theta \, d\theta \) is a standard integral equal to \( \ln | \sec \theta + \tan \theta | + C \).

Step 6: Back-substitution

Since \( x = \frac{3}{5} \sec \theta \), we have \( \sec \theta = \frac{5x}{3} \) and \( \tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\left(\frac{5x}{3}\right)^2 - 1} = \sqrt{\frac{25x^2}{9} - 1} \). Simplifying, we find: \[ \tan \theta = \frac{\sqrt{25x^2 - 9}}{3} \] Substitute back to get the original variable \( x \) back in terms of \( \theta \): \[ \ln \left| \frac{5x}{3} + \frac{\sqrt{25x^2 - 9}}{3} \right| + C \] This simplifies to: \[ \ln \left| \frac{5x + \sqrt{25x^2 - 9}}{3} \right| + C \] Finally, factor out the constant from the logarithm: \[ \ln \left| 5x + \sqrt{25x^2 - 9} \right| - \ln 3 + C \] Which can be rewritten as: \[ \ln \left| 5x + \sqrt{25x^2 - 9} \right| + C' \] Where \( C' \) is the adjusted constant of integration.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in integral calculus to simplify integrals involving square roots of expressions like \( a^2 x^2 - b^2 \). This method leverages the identities of trigonometric functions to transform complex algebraic expressions into simpler trigonometric forms. By substituting \( x = \frac{3}{5} \sec \theta \) in the original integral \( \int \frac{5 \, dx}{\sqrt{25x^2 - 9}} \), we align the problem with trigonometric identities.

• Firstly, it eliminates the square root, simplifying the expression \( \sqrt{25x^2 - 9} \) into \( 3\tan \theta \).
• Secondly, the substitution \( dx = \frac{3}{5} \sec \theta \tan \theta \, d\theta \) modifies the differential element to fit the trigonometric context.
• Finally, integrating becomes feasible with the expression \( \int \sec \theta \, d\theta \).

Trigonometric substitutions not only simplify the integration process but also enhance comprehension of how trigonometric relationships correlate with algebraic forms. This method is particularly useful when dealing with expressions that resemble the Pythagorean identity \( \sec^2 \theta - 1 = \tan^2 \theta \).

Indefinite Integrals

Indefinite integrals represent the reverse process of differentiation. When you solve an indefinite integral, you seek the antiderivative plus a constant of integration, denoted by \( C \). In the exercise at hand, the expression \( \int \frac{5 \, dx}{\sqrt{25x^2 - 9}} \) is evaluated to become \( \ln \left| 5x + \sqrt{25x^2 - 9} \right| + C' \).

• The concept of indefinite integrals is crucial in finding general solutions to functions without specified limits.
• These solutions include a constant \( C \) because differentiation of a constant always results in zero, thus the specific value of \( C \) could be any constant.
• The purpose of the constant is to account for all possible shifts of the antiderivative along the y-axis.

Understanding indefinite integrals is key not just in solving calculus problems, but in capturing the broader motion and behavior of functions. As with the example provided, the integral is solved in such a way that it demonstrates how transformations using trigonometric identities can aid in evaluating indefinite integrals.

Integration Techniques

When dealing with complex integrals, such as those including square roots or trigonometric forms, applying effective integration techniques is vital. The practice of integral calculus involves several strategies, each suited for different kinds of functions. Here, the technique of trigonometric substitution was effective for simplifying the integral \( \int \sec \theta \, d\theta \).

• Substitution—using methods like \( x = \tan \theta \) or \( x = \sec \theta \)—reduces complex expressions into manageable trigonometric integrals.
• Recognizing standard forms and identities simplifies the integration process, saving time and effort.
• Solutions require checking back-substitution validity, ensuring the transformation doesn't affect the limits or the function outside the interval of integration.

Mastering varying integration techniques, such as substitution, partial fractions, or integration by parts, provides a robust toolkit for handling diverse and challenging calculus problems. This particular exercise highlights the importance of trigonometric substitution, enabling the breakdown of seemingly difficult integrals into practicable forms.