Question

Determine the following indefinite integrals. \(\int \frac{d x}{x \sqrt{16+x^{2}}}\)

Short answer

Question: Find the indefinite integral of the function \(\int \frac{dx}{x\sqrt{16+x^2}}\). Answer: \(\int \frac{dx}{x\sqrt{16+x^2}} = \frac{1}{4}(\ln\left|\frac{x}{4}\right| + C)\)

Step-by-step solution

Identify the trigonometric substitution

We will use the trigonometric substitution \(x = 4\tan(\theta)\), since this results in a simpler integral and \(x^2\) in the denominator becomes \(16\tan^2(\theta)\).

Differentiate the substitution

Now, let's find the differential of our substitution: \(dx = 4\sec^2(\theta) d\theta\).

Substitute back into the integral

Substituting both \(x\) and \(dx\) into the integral, we get: \(\int \frac{4\sec^2(\theta) d\theta}{4\tan(\theta) \sqrt{16+ 16\tan^2(\theta)}}\).

Simplify the integral

Simplify the integral as follows: \(\int \frac{\sec^2(\theta) d\theta}{\tan(\theta) \sqrt{16\sec^2(\theta)}}\) Using the trigonometric identity, \(\sec^2(\theta)=1+\tan^2(\theta)\), we have, \(\int \frac{\sec^2(\theta) d\theta}{\tan(\theta)\cdot4\sec(\theta)}\) Now, cancel out the common term \(\sec(\theta)\), and we get, \(\frac{1}{4}\int \frac{\sec(\theta)d\theta}{\tan(\theta)}\)

Perform a new substitution

Let's set \(u=\tan(\theta)\). Then, \(du=\sec^2(\theta) d\theta\). Substituting \(u\) and \(du\) back into the integral, we get: \(\frac{1}{4}\int \frac{du}{u}\)

Integrate and substitute back

Now, integrating with respect to \(u\), we have: \(\frac{1}{4}(\ln|u| + C)\) Substituting back \(u = \tan(\theta)\) gives: \(\frac{1}{4}(\ln|\tan(\theta)| + C)\) Now, we need to replace \(\theta\) with our original variable \(x\). Recall that we made the substitution \(x = 4\tan(\theta)\), so \(\tan(\theta) = \frac{x}{4}\).

Write the final answer

Replace \(\tan(\theta)\) with \(\frac{x}{4}\) in the expression: \(\frac{1}{4}(\ln\left|\frac{x}{4}\right| + C)\) This is the final answer for the indefinite integral of the given function: \(\int \frac{dx}{x\sqrt{16+x^2}} = \frac{1}{4}(\ln\left|\frac{x}{4}\right| + C)\)

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in calculus to evaluate integrals that involve square roots of quadratic expressions. The main idea is to substitute a trigonometric function for a variable to simplify the integral into a more manageable form.
Specifically, for expressions like \ \( a^2 + x^2 \ \), you can use substitutions like \ \( x = a \tan(\theta) \ \). This helps because it leverages trigonometric identities to transform complicated square roots into simpler expressions. In our exercise, we used \ \( x = 4 \tan(\theta) \ \) since the denominator included \ \( \sqrt{16 + x^2} \ \). This allowed us to use the identity \ \( 1 + \tan^2(\theta) = \sec^2(\theta) \ \) to simplify the integrand.
As a result, the square root \ \( \sqrt{16 + 16\tan^2(\theta)} \ \) effortlessly simplifies to \ \( 4\sec(\theta) \ \), making the overall integration process much smoother.

Integration Techniques

Integration techniques refer to various methods and strategies used to solve integrals. When faced with complex integrals, you can employ specific techniques to convert them into simpler forms.
Techniques like substitution, integration by parts, and trigonometric substitution are common among these methods.
In the given problem, the initial step was to apply trigonometric substitution. But after the first substitution, a new substitution using \ \( u = \tan(\theta) \ \) was also required, simplifying the integral to \ \( \int \frac{du}{u} \ \).
This resulted in a \ \( \ln(|u|) \ \) form, which is a basic natural logarithm integral.

• Using a chain of substitutions can be crucial when dealing with complex integrals.
• Simplifying intermediates is fundamental to achieving the correct answer efficiently and accurately.
• Recognizing opportunities to utilize logarithmic form in integrals can often dramatically simplify the problem.

Calculus

Calculus is a branch of mathematics focused on change and motion, encompassing derivatives and integrals. While derivatives focus on the rate of change, integrals deal with accumulation. The concept of integrals is often applied to find areas under curves, among other applications.

Indefinite integrals, like the one in this problem, represent families of functions and include an arbitrary constant \ \( C \ \). The method tackles finding the antiderivative or primitive of the function. Presenting solutions, like in our problem with trigonometric substitution, showcases how calculus merges different mathematical techniques to solve complex equations.

Understanding calculus requires a strong grasp of fundamental principles:

• Recognize the types of functions that can utilize trigonometric identities.
• Practice changing variables and simplifying expressions, which is essential for managing integrals.
• Apply substitution strategies accordingly, paying close attention to differentiating appropriately.

This ensures seamless transitions between representations, making it easier to apply theory to practice.