Question

Integrals Involving Inverse Hyperbolic Functions Evaluate the following integrals: a. \(\int \frac{1}{\sqrt{4 x^{2}-1}} d x\) b. \(\int \frac{1}{2 x \sqrt{1-9 x^{2}}} d x\)

Short answer

a: \( \frac{1}{2} \sinh^{-1}(2x) + C \); b: \( \frac{1}{3} \cosh^{-1}(3x) + C \)."

Step-by-step solution

Recognize the form of inverse hyperbolic functions

For integral a, the integrand resembles the derivative of the inverse hyperbolic sine function, \( \sinh^{-1} x = \int \frac{1}{\sqrt{x^2 + 1}} \, dx \). We need to manipulate the integrand to this form.

Simplification using substitution (Integral a)

Given integrand is \( \frac{1}{\sqrt{4x^2 - 1}} \). Set \( u = 2x \), then \( du = 2 dx \) or \( dx = \frac{du}{2} \). The integral becomes \( \int \frac{1}{\sqrt{u^2 - 1}} \cdot \frac{1}{2} du \), which is \( \frac{1}{2} \int \frac{1}{\sqrt{u^2 - 1}} \, du \).

Use known integral for arcsinh (Integral a)

The integral \( \int \frac{1}{\sqrt{u^2 - 1}} \, du \) is known as \( \sinh^{-1} u \). So, \( \frac{1}{2} \int \frac{1}{\sqrt{u^2 - 1}} \, du = \frac{1}{2} \sinh^{-1} u + C \). Substitute back \( u = 2x \), so the result is \( \frac{1}{2} \sinh^{-1}(2x) + C \).

Recognize form and substitution for Integral b

The integrand \( \frac{1}{2x\sqrt{1-9x^2}} \) suggests an inverse hyperbolic sine substitution, where \( \cosh^{-1}(x) = \int \frac{1}{x\sqrt{x^2-1}} \, dx \). Set \( u = 3x \), then \( du = 3 \, dx \) or \( dx = \frac{du}{3} \). The integral becomes \( \frac{1}{6} \int \frac{1}{u\sqrt{1-u^2/9}} \, du \).

Simplifying Integral b

Rewrite the integrand as \( \frac{1/3}{u\sqrt{1-\frac{u^2}{9}}} \) and notice it simplifies to \( \frac{1}{u\sqrt{1-u^2}} \times \frac{1}{3} \). This matches \( \cosh^{-1}(u) \) scaled by \( \frac{1}{3} \).

Evaluate the Integral b

The integral \( \int \frac{1}{u\sqrt{1-u^2}} \, du = \cosh^{-1}(u) \), so \( \frac{1}{3} \int \frac{1}{u\sqrt{1-u^2}} \, du = \frac{1}{3} \cosh^{-1}(u) + C \). Substitute back \( u = 3x \), so the result is \( \frac{1}{3} \cosh^{-1}(3x) + C \).

Key concepts

Integration Techniques

Integration is all about finding the original function from its derivative. When working with integrals, there are various techniques that help us solve them effectively. Learning these integration techniques is crucial in calculus. They give us powerful tools to approach problems that might seem difficult at first glance.

One important technique is recognizing the derivative's shape and comparing it with known integrals of functions, such as trigonometric or hyperbolic functions. For instance, in our exercise, we need to evaluate integrals involving inverse hyperbolic functions. By recognizing the form of the integrand, we can apply specific integration techniques to find the solution.

Another useful approach is partial fraction decomposition, which is handy when dealing with rational functions. Understanding these tools enriches our skillset, allowing us to tackle challenging calculus problems with more confidence.

Substitution Method

The substitution method, also known as u-substitution, is a technique that simplifies the process of integration by changing variables. This method is often used when dealing with complex integrals to make them easier to solve.

Here's how it works: you choose a substitution to transform the integral into a simpler form. This involves replacing a part of the integrand with a new variable, usually denoted as "u." Once the substitution is made, the integral in terms of "u" is often more straightforward.

In the exercise, we used substitution to handle integrals that involved inverse hyperbolic functions. For example, by letting \(u = 2x\), we could transform the integral \(\int \frac{1}{\sqrt{4x^2 - 1}} \; dx\) into a more manageable form \(\frac{1}{2} \int \frac{1}{\sqrt{u^2 - 1}} \; du\). This method is key to simplifying complex integral problems and was essential to finding our final solutions.

Antiderivatives

Antiderivatives, or indefinite integrals, are the functions you'd obtain when integrating another function. The antiderivative tells us what function would produce the given derivative. Finding these functions is a fundamental part of calculus and is pivotal in solving integrals.

When you find an antiderivative, you will notice a "+ C" at the end of your solution. This "C" represents the constant of integration because there are infinitely many functions whose derivative is the function you started with. They differ only by a constant. For instance, when you integrate \(\frac{1}{\sqrt{u^2 - 1}}\) to find \(\sinh^{-1}(u)\), you add "+ C" to show the family of functions that share the derivative.

Recognizing antiderivatives allows us to revert the differentiation process. This process also helps in solving initial value problems and in applying integration techniques like substitution.

Calculus

Calculus is the mathematical study of change, focusing on using limits, derivatives, integrals, and infinite series. It provides the tools to describe and understand how things change and move and is involved in various applications across science and engineering.

Within calculus, integration and differentiation are two primary operations. Integration is often considered the reverse process of differentiation. It allows us to calculate things like areas under curves and cumulative quantities, such as distance traveled.

The inverse hyperbolic functions in this exercise are part of calculus's broad functions family, which helps us solve real-world problems involving exponential growth and decay or even the shapes of suspended cables. By carefully selecting integration techniques, such as substitution, and identifying antiderivatives, calculus enables the solving of complex mathematical problems efficiently.