Question

In the following exercises, evaluate each integral in terms of an inverse trigonometric function. \(\int_{-1 / 2}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}}\)

Short answer

The integral evaluates to \( \frac{\pi}{3} \).

Step-by-step solution

Identify the Integral Form

Recognize that the integral \[ \int \frac{1} {\sqrt{1-x^2}} \, dx \]fits the form of the derivative of the inverse sine function: \[ \frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \].Therefore, the integral in terms of inverse trigonometric function is \[ \sin^{-1} x + C \],where \( C \) is the constant of integration.

Evaluate the Definite Integral

Since the integral is definite from \(-\frac{1}{2}\) to \(\frac{1}{2}\),we need to calculate \[ \sin^{-1} \left(\frac{1}{2}\right) - \sin^{-1} \left(-\frac{1}{2}\right) \].We utilize the symmetry and properties of inverse sine function: \( \sin^{-1} \left(-\frac{1}{2}\right) = -\sin^{-1} \left(\frac{1}{2}\right) \).

Simplify the Expression

Insert the value for \( \sin^{-1} \left(\frac{1}{2}\right) \),which is \( \frac{\pi}{6} \).Thus,\[ \sin^{-1} \left(\frac{1}{2}\right) - (-\sin^{-1} \left(\frac{1}{2}\right)) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \].

Key concepts

Definite Integrals

A definite integral is a way of calculating the area under a curve between two specified points. It's like slicing up a graph and adding up all the bits to find the total area. For example, in this exercise, we find the area under the curve of \( \int_{-1 / 2}^{1 / 2} \frac{1}{\sqrt{1-x^{2}}} \, dx \) between \(-\frac{1}{2}\) and \(\frac{1}{2}\).

• The lower bound of the integral is \(-\frac{1}{2}\), and the upper bound is \(\frac{1}{2}\).
• The limits of integration are essential because they specify the portion of the curve we’re interested in.

The result provides the accumulated "net area" between the curve and the x-axis, accounting for any portions above and below the x-axis. This practice is crucial in applications like physics and engineering to find quantities such as displacement and total work done.

Trigonometric Integration

Trigonometric integration involves finding integrals that include trigonometric functions like sine, cosine, and tangent. In particular, when dealing with integrals that relate to inverse trig functions, we often look for the matching derivative.
In this exercise, the integral \( \int \frac{1} {\sqrt{1-x^2}} \, dx \) matches the derivative of the inverse sine function \( \sin^{-1} x \). Understanding how these trigonometric integrals connect to their derivatives is crucial in evaluating them.

• Recognizing patterns is essential. For this integral, the pattern belongs to the arcsine function's derivative.
• Derivatives provide the key to "reversing" or solving the integral.

This approach of pattern recognition helps solve integrals that initially seem complex and direct them to a simpler algebraic form.

Integral Evaluation

Integral evaluation means unraveling an integral to find its value or a function's area representation. There are primarily two types: indefinite and definite integrals. For definite integrals, like the one in this problem, we focus on calculating exact values. Here’s how it’s done:
After recognizing the integral as \( \sin^{-1} x \), we calculated it over the given bounds. We solved for \( \sin^{-1} \left(\frac{1}{2}\right) - \sin^{-1} \left(-\frac{1}{2}\right) \), which means evaluating the function at these points.

• First, compute the antiderivative if needed.
• Then, substitute the upper boundary and lower boundary into the antiderivative.
• Finally, subtract these values to get the result.

Integrating correctly gives us the area under the specified function's curve between bounds, a fundamental skill in calculus.

Arcsine Function

The arcsine function, noted as \( \sin^{-1} x \), is the inverse of the sine function. It is a vital tool in trigonometry. In calculus, it helps to express integrals involving forms like \( \frac{1}{\sqrt{1-x^2}} \). Here’s how it works:

• Arcsine is the angle whose sine is a given number. If \( \sin y = x \), then \( y = \sin^{-1} x \).
• The usual range for the arcsine function is \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), making the function invertible.

Within this exercise, solving \( \sin^{-1} \left(\frac{1}{2}\right) \) involves finding the angle whose sine is \(\frac{1}{2}\), known to be \( \frac{\pi}{6} \). It is special due to its utility in problems involving inverse trigonometric forms, offering a path from an integral back to a function.