Question

Evaluate the improper integrals. Each of these integrals has an infinite discontinuity either at an endpoint or at an interior point of the interval.\(\int_{0}^{3} \frac{d x}{\sqrt{9-x^{2}}}\)

Short answer

The integral evaluates to \( \frac{\pi}{2} \).

Step-by-step solution

Recognize the Form of the Integral

The integral \( \int \frac{dx}{\sqrt{9-x^2}} \) suggests a trigonometric substitution because it resembles the derivative of an inverse trigonometric function.

Substitution Using Trigonometric Functions

Set \( x = 3\sin\theta \). This gives \( dx = 3\cos\theta\,d\theta \) and transforms the limits of integration: when \( x = 0 \), \( \theta = 0 \); when \( x = 3 \), \( \theta = \frac{\pi}{2} \). The integral becomes \( \int_0^{\frac{\pi}{2}} \frac{3\cos\theta\,d\theta}{\sqrt{9-(3\sin\theta)^2}} \).

Simplify the Integral

Simplifying inside the integral, we have \( \sqrt{9-9\sin^2\theta} = \sqrt{9\cos^2\theta} = 3\cos\theta \). This simplifies the integral to \( \int_0^{\frac{\pi}{2}} d\theta \).

Evaluate the Simplified Integral

The integral \( \int_0^{\frac{\pi}{2}} d\theta \) is straightforward to evaluate. It yields \([\theta]_0^{\frac{\pi}{2}} = \theta\big|_0^{\frac{\pi}{2}} = \frac{\pi}{2} - 0\).

Conclude with the Final Result

The evaluated integral gives \( \frac{\pi}{2} \), indicating that the improper integral converges to this value.

Key concepts

Trigonometric Substitution

When faced with integrals that involve expressions like \( \sqrt{a^2 - x^2} \), trigonometric substitution may provide a clearer path to solving them. This technique transforms the integral into a more familiar form by substituting a variable that utilizes trigonometric identities.
In this specific exercise, the expression \( \sqrt{9-x^2} \) resembles the derivative of an inverse trigonometric function. By setting \( x = 3\sin\theta \), the variable change takes advantage of the Pythagorean identity: \( \sin^2\theta + \cos^2\theta = 1 \). This helps simplify the integrand since:

• \(1 - \sin^2\theta = \cos^2\theta\), which allows us to express the square root as 3\cos\theta.
• Substitution also transforms the differential \( dx \) into \( 3\cos\theta\,d\theta \).

Subsequently, the limits of integration are converted using the relationship \( x = 3\sin\theta \), causing them to shift from \( x = 0 \) to \( 3 \), into \( \theta = 0 \) to \( \frac{\pi}{2} \). This simplification results in a much easier integral to evaluate.

Infinite Discontinuity

Improper integrals often involve infinite discontinuities either at the integrand's endpoints or within its range. These discontinuities complicate direct evaluation because the function may not be well-behaved, specifically infinite or undefined at certain points.
In the exercise at hand, there's a potential infinite discontinuity at the endpoint \( x = 3 \) because \( \sqrt{9 - x^2} \) approaches zero, leading to division by zero and thus making the integrand undefined.

• To handle this, trigonometric substitution provides an angle-based framework, transforming the function to avoid direct confrontation with the discontinuity by making the operations in terms of \( \theta \).

The process essentially redefines the problematic boundaries in terms of the trigonometric range, allowing the problematic point to be dealt with and the integral evaluated despite the infinite discontinuity.

Convergence of Integrals

Convergence refers to the ability of the integral to approach a finite limit. In the context of improper integrals, convergence is pivotal since these integrals often extend over infinite intervals or involve functions with unbounded behavior.
In this problem, using trigonometric substitution streamlined the integral to the form \( \int_0^{\frac{\pi}{2}} d\theta \), leading to a definitive result of \( \frac{\pi}{2} \). This outcome signifies that the improper integral converges, meaning the area under the curve over the given range is finite.
Ensuring convergence can be critical in higher mathematics, where integrals of divergent functions lead to no solution or conceptual paradoxes. The finality of the result \( \frac{\pi}{2} \) confirms stability and predictability in the solution, validating the method employed to reach it. This underscores the importance of substitution and understanding discontinuities for confirming whether an improper integral converges or diverges.