Question

In the following exercises, evaluate each integral in terms of an inverse trigonometric function. \(\int_{0}^{\sqrt{3} / 2} \frac{d x}{\sqrt{1-x^{2}}}\) =

Short answer

The integral evaluates to \( \pi/3 \).

Step-by-step solution

Identify the Integral Type

The given integral is \( \int \frac{1}{\sqrt{1-x^2}} \, dx \). This resembles the derivative of the inverse sine function, which is \( \frac{d}{dx}[ \sin^{-1}(x) ] = \frac{1}{\sqrt{1-x^2}} \). Thus, we recognize this as an integral that can be expressed in terms of \( \sin^{-1}(x) \).

Apply the Antiderivative Formula

Based on the recognition, the antiderivative of \( \frac{1}{\sqrt{1-x^2}} \) is \( \sin^{-1}(x) + C \). So, the indefinite integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) is \( \sin^{-1}(x) + C \).

Evaluate the Definite Integral

The definite integral \( \int_{0}^{\sqrt{3}/2} \frac{1}{\sqrt{1-x^2}} \, dx \) can be solved using the antiderivative \( \sin^{-1}(x) \) evaluated from \( 0 \) to \( \sqrt{3}/2 \). This gives us \( \sin^{-1}(\sqrt{3}/2) - \sin^{-1}(0) \).

Calculate the Trigonometric Values

Calculate \( \sin^{-1}(\sqrt{3}/2) \). Since \( \sin(\pi/3) = \sqrt{3}/2 \), it follows that \( \sin^{-1}(\sqrt{3}/2) = \pi/3 \). Also, \( \sin^{-1}(0) = 0 \).

Find the Final Result

Using these values, compute \( \sin^{-1}(\sqrt{3}/2) - \sin^{-1}(0) = \pi/3 - 0 = \pi/3 \). The final result is \( \pi/3 \).

Key concepts

Integral Evaluation

Evaluating an integral involves finding the function whose derivative results in the integrand. In this exercise, we're asked to evaluate the integral \( \int_{0}^{\sqrt{3}/2} \frac{1}{\sqrt{1-x^2}} \, dx \). This integral is notable because it matches a standard form that's connected to inverse trigonometric functions. Specifically, this integral form relates to the derivative of the inverse sine function, \( \frac{d}{dx}[ \sin^{-1}(x) ] = \frac{1}{\sqrt{1-x^2}} \).

Once the type of integral is identified, you can apply this knowledge to replace the integral with its antiderivative. In this case, since the integral matches the derivative of \( \sin^{-1}(x) \), the antiderivative is simply \( \sin^{-1}(x) + C \). This understanding is crucial for evaluating definite integrals accurately, allowing you to determine their exact values between specified limits.

Antiderivatives

The concept of antiderivatives is central to solving integrals. An antiderivative of a function is another function that, when differentiated, returns the original function. In our exercise, we identified that \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) can be linked directly to the antiderivative of \( \sin^{-1}(x) \). This means that the indefinite integral, or antiderivative, is \( \sin^{-1}(x) + C \), where \( C \) represents the constant of integration.

Understanding this process simplifies evaluating definite integrals, as you can focus on calculating specific values using the antiderivative within given limits. When you solve the definite integral, you won't include \( C \), because constant terms cancel each other out during the subtraction process of evaluating from the upper to the lower limit.

Definite Integrals

A definite integral represents the accumulation of quantities over a given interval, akin to finding the area under a curve. In our exercise, the definite integral is \( \int_{0}^{\sqrt{3}/2} \frac{1}{\sqrt{1-x^2}} \, dx \). Once we've found the antiderivative \( \sin^{-1}(x) \), the task becomes one of evaluating this expression at the boundaries of the integral.

To compute this, you calculate \( \sin^{-1}(\sqrt{3}/2) \) and \( \sin^{-1}(0) \). Knowing the values of these inverse trigonometric functions simplifies the computation. For example, \( \sin^{-1}(\sqrt{3}/2) = \pi/3 \) because \( \sin(\pi/3) = \sqrt{3}/2 \). Similarly, \( \sin^{-1}(0) = 0 \). So the definite integral evaluates to \( \pi/3 - 0 = \pi/3 \).

This illustrates not only the process of using antiderivatives but also how definite integrals condense wide-ranging information into a single, concise value.