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Chapter 7: Problem 40

Evaluate the following integrals. $$\int \frac{d x}{x^{2} \sqrt{9 x^{2}-1}}, x>\frac{1}{3}$$

Short Answer

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Question: Evaluate the integral $$\int \frac{d x}{x^{2} \sqrt{9x^{2}-1}}, x>\frac{1}{3}$$ Answer: The integral evaluates to $$\int \frac{d x}{x^{2} \sqrt{9 x^{2}-1}} = \frac{1}{2}\cdot\frac{\sqrt{9x^2-1}}{x} + C$$

Step by step solution

01

Determine the appropriate substitution

We notice that the integral has a term of the form \(\sqrt{9x^{2}-1}\). We can choose a trigonometric substitution such that this square root term simplifies to a simpler expression. We let $$x = \frac{1}{3}\sec\theta$$ Differentiating both sides with respect to \(\theta\), we have $$\frac{d x}{d \theta} = \frac{1}{3}\sec\theta\tan\theta$$ Now, substitute the expression for \(x\) into the square root term: $$\sqrt{9x^{2}-1} = \sqrt{9\left(\frac{1}{3}\sec\theta\right)^{2}-1} = \frac{1}{3}\sqrt{9\sec^2\theta - 1}$$ Using the identity \(\sec^2\theta = \tan^2\theta + 1\), we simplify the square root term: $$\frac{1}{3}\sqrt{9\sec^2\theta - 1} = \frac{1}{3}\sqrt{9(\tan^2\theta + 1)-1} = \frac{1}{3}\sqrt{9\tan^2\theta + 8} = \frac{2}{3}|\tan\theta|$$
02

Substitute and simplify the integral

Now, substitute \(x = \frac{1}{3}\sec\theta\) and \(d x = \frac{1}{3}\sec\theta\tan\theta d \theta\) into the original integral. Note that since \(x > \frac{1}{3}\), we have \(|\tan\theta| = \tan\theta\). $$\int \frac{d x}{x^{2} \sqrt{9x^{2}-1}} = \int \frac{\frac{1}{3}\sec\theta\tan\theta d \theta}{\left(\frac{1}{3}\sec\theta\right)^{2}\cdot \frac{2}{3}\tan\theta} = 3\int \frac{\sec\theta \tan\theta d \theta}{2\sec^2\theta\tan\theta} = \frac{3}{2} \int \frac{d \theta}{\sec\theta}$$
03

Integrate the simplified expression

Now, we need to integrate the expression \(\frac{3}{2}\int\frac{d\theta}{\sec\theta}\). This integral can be evaluated using the identity \(1 = \sin^2\theta + \cos^2\theta\). $$\frac{3}{2}\int\frac{d\theta}{\sec\theta} = \frac{3}{2}\int\frac{\cos{\theta}}{1}d\theta = \frac{3}{2}\int \cos{\theta}d\theta$$ Now integrate with respect to \(\theta\): $$\frac{3}{2}\int \cos{\theta}d\theta = \frac{3}{2}\sin{\theta} + C$$
04

Substitute back for x

Now, we need to rewrite the result in terms of \(x\). Recall that \(x = \frac{1}{3}\sec\theta\), which implies \(\theta = \sec^{-1}(3x)\). Using the identity \(\sin^2\theta = 1 - \cos^2\theta\), we can find that \(\sin\theta = \sqrt{1 - \cos^2\theta}\). Using the Pythagorean identity \(\cos\theta = \frac{1}{\sec\theta}\), we get $$\sin\theta = \sqrt{1-\left(\frac{1}{3x}\right)^2} = \frac{\sqrt{9x^2-1}}{3x}$$ Finally, substitute this back into our integral: $$\frac{3}{2}\sin{\theta} + C = \frac{3}{2}\cdot\frac{\sqrt{9x^2-1}}{3x} + C = \frac{1}{2}\cdot\frac{\sqrt{9x^2-1}}{x} + C$$ So the integral evaluates to: $$\int \frac{d x}{x^{2} \sqrt{9 x^{2}-1}} = \frac{1}{2}\cdot\frac{\sqrt{9x^2-1}}{x} + C$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Substitution
Trigonometric substitution is a powerful technique in integration that simplifies expressions by turning them into familiar trigonometric identities. It's particularly useful for integrating functions involving square roots like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). In such situations, we substitute a trigonometric function for \( x \) to exploit the Pythagorean identities.

In this particular problem, the integrand involves \( \sqrt{9x^2 - 1} \), suggesting the use of the secant function. The substitution \( x = \frac{1}{3}\sec\theta \) is chosen because it transforms the square root using the identity \( \sec^2\theta = 1 + \tan^2\theta \), effectively simplifying the integral.

This method turns complex algebraic expressions into simpler trigonometric forms, making subsequent steps easier to handle. Hence, trigonometric substitution is crucial when dealing with similar integrals involving square roots. It's all about choosing the right trigonometric substitution to match the form of the square root expression.
Definite Integrals
Definite integrals provide the accumulated value of a function over a specific interval, contrasting with indefinite integrals, which include a constant of integration. Through definite integrals, we determine areas under curves, regardless of the function spanning above or below the x-axis.

In this exercise, although the focus is on evaluating an indefinite integral, the concept of substitution techniques, such as trigonometric substitution, is equally applicable to definite integrals. When solving definite integrals with substitution, the limits of integration must also be transformed according to the substitution employed. This ensures that the new bounds align with the variable change, preserving the integral's original domain.

Understanding how to adjust the limits of integration or interpret results in terms of the original variable is fundamental in applying these techniques to definite integrals. This ensures accurate calculations and interpretations of the integral's numerical results.
Pythagorean Identity
The Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \) is a cornerstone in trigonometry, frequently employed to simplify integrals. This identity extends to other forms, such as \( \sec^2\theta = 1 + \tan^2\theta \), which are handy for trigonometric substitution in integration problems.

In the worked solution, the expression \( \sqrt{9x^2 - 1} \) is simplified using the identity, where \( \sec\theta \) and \( \tan\theta \) are related via \( \sec^2\theta = 1 + \tan^2\theta \). Consequently, the square root simplifies to a multiple of \( \tan\theta \), further easing the integration task.

By understanding and applying Pythagorean identities, complex expressions are translated into trigonometric functions, providing a pathway to simpler computations. This technique emphasizes the interconnection between algebraic expressions and trigonometric identities, illustrating the power of these fundamental concepts in calculus integration.

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