Question

Evaluate the integrals. If the integral diverges, answer "diverges."\(\int_{-1}^{1} \frac{d x}{\sqrt{1-x^{2}}}\)

Short answer

The value of the integral is \(\pi\).

Step-by-step solution

Identify the Type of Integral

The integral \(\int_{-1}^{1} \frac{dx}{\sqrt{1-x^2}}\) is a definite integral with symmetric limits. Notice that the integrand is \(\frac{1}{\sqrt{1-x^2}}\), which resembles the standard trigonometric identity \(\sin^2(\theta) + \cos^2(\theta) = 1\). Thus, this integral represents an inverse trigonometric function.

Use Trigonometric Substitution

To solve this integral, use the substitution \(x = \sin(\theta)\). Consequently, \(\sqrt{1-x^2} = \cos(\theta)\) and \(dx = \cos(\theta)d\theta\). This substitution converts the integral into \(\int \frac{\cos(\theta)d\theta}{\cos(\theta)}\), which simplifies to \(\int d\theta\).

Change the Limits of Integration

When \(x = -1\), \(\sin(\theta) = -1\) implies \(\theta = -\frac{\pi}{2}\). When \(x = 1\), \(\sin(\theta) = 1\) implies \(\theta = \frac{\pi}{2}\). Therefore, the integral bounds change from \([-1, 1]\) to \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

Evaluate the Integral

The integral \(\int d\theta\) from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) results in \([\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \left[\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right] = \pi\).

Check for Convergence

Although the integral bounds are near discontinuities where \(x = \pm1\), the integral converges as the function remains bounded within the limits of integration. Thus, there is no issue of divergence in this case.

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique to simplify integrals involving square roots, especially those resembling the Pythagorean identities. In the case of the integral \(\int \frac{dx}{\sqrt{1-x^2}}\), the expression \(\sqrt{1-x^2}\) suggests a connection to the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\). Thus, substituting \(x = \sin(\theta)\) is a logical choice.
This substitution transforms the troublesome square root \(\sqrt{1-x^2}\) into \(\cos(\theta)\), significantly simplifying the problem.
Steps to perform trigonometric substitution:

• Choose a substitution that simplifies the integral. For \(\sqrt{1-x^2}\), use \(x = \sin(\theta)\).
• Find the differential \(dx\) in terms of \(d\theta\). In this case, \(dx = \cos(\theta)d\theta\).
• Convert the integral into one involving \(\theta\), which usually reduces complexity.

This technique can make evaluating definite integrals easier, especially when the integrand matches a known trigonometric identity.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as arcsine, arise in integrals involving specific algebraic expressions. In our exercise, the integral \(\int \frac{dx}{\sqrt{1-x^2}}\) is a classic example that leads to arc sine. The substitution \(x = \sin(\theta)\) directly indicates an inverse trigonometric relationship.
The key functions include:

• \(\arcsin(x)\)
• \(\arccos(x)\)
• \(\arctan(x)\)

These functions allow us to solve integrals that cannot be easily expressed in terms of basic algebraic operations.
By transforming the original integral into \(\int d\theta\), which simplifies to computing the angular range from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), the solution leverages the inverse relationships. The result \(\pi\) showcases the symmetry and simplicity of inverse trigonometric integrals when approached with substitution.

Convergence of Integrals

When dealing with definite integrals, especially those reaching limits where the function might become undefined, assessing convergence is crucial. In our integral \(\int_{-1}^{1} \frac{dx}{\sqrt{1-x^2}}\), potential issues arise due to the bounds at \(-1\) and \(1\). Here, the function looks like it could become undefined as \(x = \pm1\) turns the square root into zero, risking a division by zero situation.
To ensure convergence, we confirm that the value of the integral isn’t infinity and remains bounded within the limits.
Checkpoints for convergence:

• Examine the behavior of the integrand near potential discontinuities or bounds.
• Assure that the result of the integral is finite.

In this case, despite potential concerns at \(x = -1\) and \(x = 1\), the integral converges due to the behavior of \(\frac{1}{\sqrt{1-x^2}}\), which remains bounded and integrable over the interval \([-1, 1]\). As confirmed by the calculation, the integral results in the finite value \(\pi\), indicating successful convergence.