Problem 78 - Thomas Calculus

Question

Evaluate each integral in Exercises \(71-82\) by using any technique you think is appropriate. $$ \int \frac{d x}{x \sqrt{4 x^{2}-1}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sec^{-1}(2x) + C \).

Step 1: Recognize the Form

The given integral is \( \int \frac{d x}{x \sqrt{4x^2 - 1}} \). This is a trigonometric integral involving \( \sqrt{a^2x^2 - b^2} \). This suggests the use of a trigonometric substitution.

Step 2: Choose the Right Substitution

For integrals involving expressions like \( \sqrt{a^2x^2 - b^2} \), a suitable substitution is \( x = \frac{1}{2} \sec(\theta) \). Then, \( dx = \frac{1}{2} \sec(\theta) \tan(\theta) \, d\theta \).

Step 3: Substitute and Simplify

Substitute \( x = \frac{1}{2} \sec(\theta) \) into the integral:\[\int \frac{\frac{1}{2} \sec(\theta) \tan(\theta) \, d\theta}{\frac{1}{2} \sec(\theta) \sqrt{4 \left(\frac{1}{2} \sec(\theta)\right)^2 - 1}}\]This simplifies to:\[\int \frac{\frac{1}{2} \sec(\theta) \tan(\theta) \, d\theta}{\frac{1}{2} \sec(\theta) \sqrt{\sec^2(\theta) - 1}}\]Note that \(\sqrt{\sec^2(\theta) - 1} = \tan(\theta)\), simplifying further.

Step 4: Simplify Further

Continuing from the substitution:\[\int \frac{\frac{1}{2} \sec(\theta) \tan(\theta)}{\frac{1}{2} \sec(\theta) \tan(\theta)} \, d\theta = \int d\theta\]Since the fraction simplifies to 1, the integral becomes \( \int d\theta = \theta + C \).

Step 5: Back Substitute

Since \( x = \frac{1}{2} \sec(\theta) \), we have \( \theta = \sec^{-1}(2x) \). Thus, the integral evaluates to:\[ \theta + C = \sec^{-1}(2x) + C\]

Step 6: Conclusion

The integral \( \int \frac{d x}{x \sqrt{4x^2 - 1}} \) evaluates to \( \sec^{-1}(2x) + C \).

Key Concepts

Integral Calculus

Integral Calculus is one of the two main branches of calculus, with the other being Differential Calculus. This branch focuses on the concept of integration, which is the mathematical process of determining the area under the curve represented by a function. Think of it as the reverse operation of differentiation. Where differentiation is concerned with rates of change, integration finds out how much something accumulates over an interval. In practical terms:

• Integration helps us calculate areas under curves and can even extend to calculate volumes.
• Solves problems related to displacement and accumulation.
• Fundamental in solving equations that describe physical, economic, and geometrical relationships.

Understanding integration begins with simple concepts like finding the net area under a curve by decomposing it into infinitesimally small parts. Advanced concepts like the Fundamental Theorem of Calculus link derivatives and integrals, showing how they are inverse processes.

Integration Techniques

Integration techniques are methods used to evaluate integrals, especially those not solvable by simple formulas. Some of the most commonly used techniques include substitution, integration by parts, partial fractions, and trigonometric substitution. Choosing the right technique:

• Substitution: Useful when the integral contains a function and its derivative. Simplifies complex expressions into easier forms.
• Integration by Parts: Applies to products of functions. It uses the integration formula \( \int u \, dv = uv - \int v \, du \).
• Trigonometric Substitution: Best used for integrals involving radical expressions of the forms \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). Such integrals often simplify when using trig functions like sine, cosine, or secant.
• Partial Fractions: Effective for rational functions. Breaks complex fractions into simpler parts for easier integration.

These techniques expand our ability to solve more complex integrals and broaden our understanding of calculus.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse functions of the basic trigonometric functions: sine, cosine, and tangent. For integration purposes, they are often used to back-substitute when trigonometric identities are used to solve integrals.Key points about inverse trigonometric functions:

• They help express angles when given the ratio of sides of a right triangle.
• Noted as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), \( \tan^{-1}(x) \), and similar for other trigonometric functions.
• In the context of integration, if a trigonometric substitution is used (e.g., \( x = \frac{1}{2} \sec(\theta) \)), returning to the variable \( x \) involves using these inverse functions.

In the original problem, after substituting with a trigonometric identity and simplifying, the solution was back-substituted using \( \sec^{-1}(2x) \), showcasing the role of inverse trigonometric functions in achieving the final answer.