Question

Evaluating a Definite Integral Using Inverse Trigonometric Functions Evaluate the definite integral \(\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}\).

Short answer

The value of the definite integral is \( \frac{\pi}{2} \).

Step-by-step solution

Identify the Integral Type

The given integral is \( \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}} \). This expression is characteristic of the inverse trigonometric function \( \arcsin(x) \), as \( \frac{d}{dx} [\arcsin(x)] = \frac{1}{\sqrt{1-x^2}} \).

Set Up the Integral Using Known Formulas

Recognize that \( \int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin(x) + C \), where \( C \) is the constant of integration. For the definite integral, we evaluate the antiderivative at the upper boundary and subtract the evaluation at the lower boundary.

Apply the Fundamental Theorem of Calculus

Use the fundamental theorem of calculus to evaluate the definite integral:\[ \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}} = \arcsin(1) - \arcsin(0) \].

Evaluate the Arc Sine Values

Evaluate \( \arcsin(1) \), which is \( \frac{\pi}{2} \), as sin(\( \frac{\pi}{2} \)) = 1.Evaluate \( \arcsin(0) \), which is 0 since sin(0) = 0.

Subtract the Results

Subtract the two values obtained: \( \frac{\pi}{2} - 0 \), which results in \( \frac{\pi}{2} \).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial in calculus, particularly when working with integrals that involve complex square roots. These functions, such as arc sine (or inverse sine), arc cosine, and arc tangent, allow us to solve integrals that would otherwise be inaccessible.

Inverse trigonometric functions enable us to convert trigonometric expressions into angles, which simplifies the evaluation of certain integrals. They invert the original trigonometric functions:

• Arc sine, denoted as \( \arcsin(x) \), is the inverse of sine.
• Arc cosine, \( \arccos(x) \), is the inverse of cosine.
• Arc tangent, \( \arctan(x) \), is the inverse of tangent.

A key aspect is their defined range. For example, the arc sine function maps a range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for outputs, understanding these ranges is essential for correctly evaluating definite integrals involving these functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of integral calculus, connecting differentiation with integration. It provides a practical way to evaluate definite integrals using antiderivatives.

This theorem states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) is given by:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]. This relationship is incredibly beneficial in evaluating integrals efficiently without needing complex calculations.
By finding an antiderivative, you can assess how a function accumulates over an interval. This was evident in our solved problem when we utilized the antiderivative \( \arcsin(x) \), making the calculation of the definite integral straightforward and precise.

Arc Sine

Arc sine, or \( \arcsin(x) \), is a specific example of an inverse trigonometric function that is particularly useful for solving integrals like our given problem. The function provides the angle whose sine is \( x \).
Its usage is clear in integrals involving expressions like \( \sqrt{1-x^2} \), where the derivative matches that of \( \arcsin(x) \):\[ \frac{d}{dx} [\arcsin(x)] = \frac{1}{\sqrt{1-x^2}} \]. This allows us to easily set up and solve definite integrals involving square roots and fractions.
Importantly, the range of \( \arcsin(x) \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), ensuring the angle calculated is valid and correctly reflects the original trigonometric relationship. By understanding how arc sine interacts with functions that match its derivative, we can streamline complex integrals effectively, as demonstrated when evaluating \( \arcsin(1) \) to \( \frac{\pi}{2} \) and \( \arcsin(0) \) to measure 0.