Question

Evaluate each integral. $$\int \frac{d x}{x \sqrt{16+x^{2}}}$$

Short answer

Question: Evaluate the integral \(\int \frac{dx}{x \sqrt{16+x^{2}}}\). Answer: The integral evaluates to \(\int \frac{dx}{x \sqrt{16+x^{2}}} = \ln|\frac{x}{4} + \sqrt{1+\frac{x^2}{16}}| + C\).

Step-by-step solution

Perform substitution

Make the substitution \(x = 4\sinh u\) to simplify the denominator. To find the differential \(dx\), differentiate \(x\) with respect to \(u\): $$\frac{dx}{du} = 4\cosh u$$ Now, solve for \(dx\): $$dx = 4\cosh u \ du$$

Replace \(x\) and \(dx\) in the integral

Substitute the expressions for \(x\) and \(dx\) back into the integral and simplify: $$\int \frac{1}{x \sqrt{16+x^{2}}} dx = \int \frac{1}{4\sinh u \cdot \sqrt{16+(4\sinh u)^{2}}} \cdot 4\cosh u \ du$$

Use hyperbolic trigonometric identity to simplify

Now, notice that we have the term \(\sqrt{16+(4\sinh u)^{2}}\) in the denominator, which can be simplified using the hyperbolic identity \(\cosh^2 u - \sinh^2 u = 1\). Rewrite the denominator using the identity: $$\sqrt{16+(4\sinh u)^{2}} = 4\cosh u$$ Now, our integral becomes: $$\int \frac{1}{4\sinh u \cdot 4\cosh u} \cdot 4\cosh u \ du$$

Simplify and integrate

Cancel the common terms and simplify the integral: $$\int \frac{1}{\sinh u} \ du$$ Now, integrate the hyperbolic function: $$\int \frac{1}{\sinh u} \ du = \ln|\sinh u + \cosh u| + C$$

Substitute back for \(x\) and simplify

Finally, substitute back the original variable \(x\) using the substitution \(x = 4\sinh u\): $$\ln|\sinh u + \cosh u| + C = \ln|\frac{x}{4} + \sqrt{1+\frac{x^2}{16}}| + C$$ The final answer is: $$\int \frac{dx}{x \sqrt{16+x^{2}}} = \ln|\frac{x}{4} + \sqrt{1+\frac{x^2}{16}}| + C$$

Key concepts

Hyperbolic Substitution

Hyperbolic substitution is a valuable technique in integral calculus, used to simplify integrals that contain square roots of expressions like \(a^2 + x^2\). The goal is to make the complicated radical disappear by leveraging hyperbolic functions. In this exercise, we used the substitution \(x = 4\sinh u\). This choice is made because hyperbolic functions have identities that mirror trigonometric identities, and \(\sinh u\) and \(\cosh u\) relate similarly to sine and cosine. This substitution transforms the original integral into a simpler form that we can more readily solve.

When applying this method:

• Replace the variable with a hyperbolic function that makes the radical manageable.
• Differentiate to find \(dx\) in terms of \(du\).
• Use the hyperbolic identity \(\cosh^2 u - \sinh^2 u = 1\) to simplify the integral further.

Switching to hyperbolic identities is particularly useful when dealing with quadratic expressions involving squares and square roots, easing the integration process.

Integration Techniques

Integration techniques are diverse, each tailor-made for simplifying different types of integrals. In this example, we employed substitution, a core technique aimed at transforming the integral into an easier form. Substitution works by changing the variable of integration to utilize identities that make the integration process straightforward.

Steps to use substitution effectively:

• Identify a possible substitution that simplifies the integral, often indicated by a troublesome radical or trigonometric expression.
• Find the differential of your substitution to allow for a complete switch of variables within the integral.
• Simplify the integral using algebra and function identities.
• Perform the integration on this simpler form, then substitute back the original variables.

Using these techniques ensures that even complex integrals can be solved by transforming them into a more familiar or manageable form.

Hyperbolic Identities

Hyperbolic identities are mathematical formulas that provide relationships between hyperbolic functions, much like the trigonometric identities do for their respective functions. They are pivotal in integration as they often simplify expressions, especially those with squares and square roots.

In this exercise, we utilized the identity \(\cosh^2 u - \sinh^2 u = 1\), which mirrors the Pythagorean identity from trigonometry. This identity allowed us to replace the unwieldy \(\sqrt{16 + (4\sinh u)^2}\) with the simpler form of \(4\cosh u\). By using these identities:

• Complex denominators can be rewritten in simpler terms.
• Facilitate cancellations that reduce the integral to an elementary form.
• Ultimately make integration direct and solvable.

Understanding these identities is essential, as they often hold the key to making certain integrals manageable. With these tools, solutions become clearer and more accessible, bridging the gap between complex expressions and straightforward integrals.