Problem 47 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{d x}{x \sqrt{25 x^{2}-2}}$$

Step-by-Step Solution

Answer

\( \ln \left| \sqrt{1 - \frac{1}{25x^2}} \right| + C \).

Step 1: Simplify the Integral

Given the integral \( \int \frac{d x}{x \sqrt{25 x^{2}-2}} \). To simplify it, notice the expression inside the square root: \( 25x^2 - 2 \). We can factor this as \( \sqrt{25(x^2 - \frac{2}{25})} \). This indicates that a substitution may simplify the integral.

Step 2: Trigonometric Substitution

Use the substitution \( x = \frac{1}{5} \sec(\theta) \). This makes \( dx = \frac{1}{5} \sec(\theta)\tan(\theta)\, d\theta \) and \( 25x^2 - 2 = 25(\frac{1}{25}\sec^2(\theta)) - 2 = \tan^2(\theta) \). The integral becomes: \[ \int \frac{\frac{1}{5} \sec(\theta)\tan(\theta)\, d\theta}{\frac{1}{5} \sec(\theta) \cdot \tan(\theta)} = \int \frac{d\theta}{\tan(\theta)}. \]

Step 3: Simplify Resulting Integral

The integral \( \int \frac{d\theta}{\tan(\theta)} \) can be rewritten using \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Therefore, \( \frac{1}{\tan(\theta)} = \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} \). The integral then simplifies to: \( \int \cot(\theta) \, d\theta \).

Step 4: Integrate \( \cot(\theta) \)

Use the identity \( \int \cot(\theta) \, d\theta = \ln |\sin(\theta)| + C \). This gives the antiderivative in terms of \( \theta \).

Step 5: Back-Substitute \( \theta \)

From the substitution \( x = \frac{1}{5} \sec(\theta) \), we have \( \sec(\theta) = 5x \) which implies \( \sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \frac{1}{25x^2}} \). Substitute back to get the result: \( \ln \left| \sqrt{1 - \frac{1}{25x^2}} \right| + C \).

Step 6: Final Solution

Apply inverse trigonometric function identities to simplify the expression further if necessary. The final integral evaluation is: \( \ln \left| \sqrt{1 - \frac{1}{25x^2}} \right| + C \).

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus, involving finding an antiderivative or calculating the area under a curve. While basic integrals can be straightforward, complex expressions often require specialized techniques for simplification. One such technique is **trigonometric substitution**, particularly useful when dealing with integrals involving square roots of quadratic expressions.
Trigonometric substitution generally involves replacing variables with trigonometric functions, such as sine, cosine, or secant. This is particularly effective because trigonometric identities simplify the algebra within the integration process. For instance, in the problem given, we used the substitution \( x = \frac{1}{5} \sec(\theta) \), transforming the original algebraic expression into a trigonometric one and significantly easing the integration process.
Understanding when and how to apply these techniques can save a lot of time and effort. It’s not just about reaching the answer, but also about simplifying the path you take to get there. Steps like identifying possible substitutions and applying them correctly are key.

Calculus Problem Solving

Successful problem solving in calculus requires a step-by-step approach. It involves understanding the problem, identifying the right technique, and then executing it with precision. In this case, solving the integral \( \int \frac{d x}{x \sqrt{25 x^{2}-2}} \) begins with examining the expression under the square root and recognizing a potential for simplification.
The process follows several crucial stages:

• Firstly, analyze the structure of the expression to predict which substitution might work best.
• Apply the chosen substitution, recalculating differential terms if necessary.
• Simplify the resulting integral, often converting to a form more manageable or recognizable for integration.
• Integrate the simplified expression, using known antiderivative forms.
• Finally, back-substitute to return to the original variable.

Iterating these steps accurately is essential. This methodology not only applies to trigonometric substitution but is a universal toolset for tackling diverse calculus problems.

Definite and Indefinite Integrals

In calculus, understanding the difference between definite and indefinite integrals is crucial. An **indefinite integral** provides a general form of antiderivatives, represented with a constant \( C \), whereas a **definite integral** calculates the exact area under a curve between two limits. The distinction lies in their applications and the necessary steps within problem-solving processes.
For indefinite integrals, like in our solved integral which resulted in a generalized antiderivative expression \( \ln \left| \sqrt{1 - \frac{1}{25x^2}} \right| + C \), there’s no specified bounds, leading to a family of potential solutions dependent on \( C \). Practically, this form is crucial for understanding the behavior or properties of functions.
Conversely, definite integrals require limits of integration and result in a numeric value, quantifying accumulated quantities such as area, displacement, etc. This makes definite integrals vital in real-world applications, including physics and engineering. Grasping these concepts equips students with the analytical skills needed to both interpret and apply calculus in theoretical and practical contexts.