Question

Evaluate \(\int_{1}^{4} \frac{d x}{\sqrt{x^{2}-1}} \cdot\) (Express the answer in exact form.)

Short answer

\(\cos^{-1}\left(\frac{1}{4}\right)\)

Step-by-step solution

Identify the Integral Type

The integral given is \[\int_{1}^{4} \frac{dx}{\sqrt{x^2-1}}.\]This integral is of the form \(\int \frac{dx}{\sqrt{x^2-a^2}}\), which is associated with inverse hyperbolic functions. Specifically, it can be evaluated using the formula for arcsecant: \[\int \frac{dx}{\sqrt{x^2-a^2}} = \text{arcsec}\left(\frac{x}{a}\right) + C.\]

Apply the Formula

In this problem, \(a = 1\). Thus, the integral results in:\[\int \frac{dx}{\sqrt{x^2-1}} = \text{arcsec}(x) + C.\] Therefore, \[\int_{1}^{4} \frac{dx}{\sqrt{x^2-1}} = \text{arcsec}(x)\bigg|_1^4.\]

Evaluate the Definite Integral

Compute the definite integral by evaluating the expression at the bounds:\[\text{arcsec}(4) - \text{arcsec}(1).\]

Calculate the Values of Arcsecant

Evaluate the value of the arcsecant function:- \(\text{arcsec}(4)\) is the angle whose secant is 4, which equals \(\cos^{-1}\left(\frac{1}{4}\right)\).- \(\text{arcsec}(1)\) is the angle whose secant is 1, which equals \(\cos^{-1}(1)\) = 0.

Subtract to Find the Result

Subtract the evaluated lower bound from the upper bound:\[\text{arcsec}(4) - \text{arcsec}(1) = \cos^{-1}\left(\frac{1}{4}\right) - 0 = \cos^{-1}\left(\frac{1}{4}\right).\] Therefore, the exact answer is \(\cos^{-1}\left(\frac{1}{4}\right)\).

Key concepts

Definite Integrals

Definite integrals provide us with a powerful tool to determine the accumulated quantity, like area under a curve, between two specific points. Unlike indefinite integrals, which yield a family of functions, definite integrals result in a single numerical value. In our exercise, the definite integral \[\int_{1}^{4} \frac{dx}{\sqrt{x^2-1}} \]was solved through specific limits from 1 to 4.

• The integral's limits (1 and 4) are crucial as they anchor the calculus operation to a particular segment of the function.
• This involves substituting the upper and lower limits into the antiderivative and then subtracting the results.

This step is essential because it translates the abstract operation of integration into a concrete numerical answer, providing insight into the behavior of functions within specific intervals.
Definite integrals can reveal properties like areas, lengths, and even probabilities depending on their applications.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are counterparts to the regular hyperbolic functions such as sinh, cosh, and tanh. They are significant in calculus for evaluating integrals that contain expressions akin to hyperbolic identities. In particular, these functions are used to solve integrals of the form:\[\int \frac{dx}{\sqrt{x^2 - a^2}}.\]

• An example of this in our exercise is recognizing the inverse hyperbolic nature of the integrand, \(\frac{1}{\sqrt{x^2-1}}\).
• This can be directly associated with the arcsecant, highlighting a unique aspect of inverse hyperbolic territories.

Inverse hyperbolic functions can often appear intimidating; however, they follow rules similar to their trigonometric counterparts, making it easier to perform manipulation using familiar calculus techniques. These functions also help in providing solutions in many real-world problems, involving high precision modeling and assessment.

Arcsecant Function

The arcsecant function, denoted as \(\text{arcsec}(x)\), is a lesser-known yet incredibly useful inverse trigonometric function. As an inverse of the secant function, it expresses angles for given secant values.
The formula used in our solution was:\[\int \frac{dx}{\sqrt{x^2-a^2}} = \text{arcsec}\left(\frac{x}{a}\right) + C.\]In our context, \(a = 1\), so the integration resolved to \(\text{arcsec}(x)\).

• The arcsecant function answers what angle has a secant equal to a certain value.
• In practical terms, it converts the challenging integration problem into an arcsecant evaluation.

When calculating arcsecant, consider its relationship with cosines because \(\text{arcsec}(x)\) computations often translate to arccosines like in our problem, where \(\text{arcsec}(4) = \cos^{-1}\left(\frac{1}{4}\right)\).
Understanding arcsecant helps decipher problems classic trigonometric functions can't manage easily. By mastering how to work with arcsecant, you expand your toolkit for solving diverse calculus problems.