Problem 50 - Calculus Single Variable

Question

In each of Exercises \(41-54,\) determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{0}^{\pi / 2} \frac{\cos (x)}{\sqrt{\sin (x)}} d x\)

Step-by-Step Solution

Answer

The integral converges and evaluates to 2.

Step 1: Understand the Improper Integral

The integral given is \(\int_{0}^{\pi / 2} \frac{\cos(x)}{\sqrt{\sin(x)}} \, dx\). It's an improper integral because the integrand \(\frac{\cos(x)}{\sqrt{\sin(x)}}\) has a problem at \(x = \frac{\pi}{2}\), where \(\sin(x)\) becomes zero, causing the denominator to approach zero.

Step 2: Evaluate the Behavior Near Problematic Point

Before integrating, evaluate the behavior of the integrand as \(x \to \frac{\pi}{2}\). Near \(\frac{\pi}{2}\), \(\sin(x) \approx 1 - (x - \frac{\pi}{2})^2/2\), meaning \(\sqrt{\sin(x)}\approx\sqrt{1-(x - \frac{\pi}{2})^2/2}\approx 1\). Therefore, \(\cos(x) \approx 0\). Thus, the integrand \(\frac{\cos(x)}{\sqrt{\sin(x)}}\) becomes zero as \(x\) approaches \(\frac{\pi}{2}\).

Step 3: Properly Handle the Improper Integral

Substitute \(y = \sin(x)\), with \(dy = \cos(x) \, dx\). The limits of integration change: when \(x = 0\), \(y = 0\) and when \(x = \frac{\pi}{2}\), \(y = 1\). The integral becomes \(\int_{0}^{1} y^{-1/2} \, dy\).

Step 4: Evaluate the New Integral

Now evaluate \(\int_{0}^{1} y^{-1/2} \, dy\). The antiderivative of \(y^{-1/2}\) is \(2y^{1/2}\). Evaluate this from \(0\) to \(1\): \(2(1^{1/2}) - 2(0^{1/2}) = 2\). Therefore, the integral converges and the value is \(2\).

Key Concepts

Convergence and Divergence

When dealing with improper integrals, one of the first steps is to determine whether the integral itself is convergent or divergent. It boils down to assessing the behavior of the integral's bounds. Specifically, we're interested in points where the integrand might become undefined or infinite.
In our exercise, the integral \( \int_{0}^{\pi / 2} \frac{\cos(x)}{\sqrt{\sin(x)}} \, dx \) is classified as improper due to the denominator \( \sqrt{\sin(x)} \) approaching zero at \( x = \frac{\pi}{2} \).
If the limit of the integral at this point shows infinity or doesn't exist, the integral is divergent. In contrast, if it approaches a finite number, it converges.
In this case, the analysis showed that the integrand effectively becomes zero as \( x \) approaches \( \frac{\pi}{2} \), suggesting potential convergence.

• Improper integrals often involve infinity as a limit or an integrand with infinite discontinuity.
• An integral's convergence is confirmed if its corresponding limit yields a finite result.
• Divergence indicates the integral approaches infinity or is undefined.

With this foundation, one can confidently determine convergence, paving the way for solution evaluation.

Antiderivatives

The term 'antiderivative' refers to a function whose derivative is equal to the given function. Finding the antiderivative is integral to solving the given problem, especially after the integral's transformation.
The original function \( \int_{0}^{\pi / 2} \frac{\cos(x)}{\sqrt{\sin(x)}} \, dx \), after substitution, became\( \int_{0}^{1} y^{-1/2} \, dy \). We sought the antiderivative of \( y^{-1/2} \).
The antiderivative of \( y^{-1/2} \) is \( 2y^{1/2} \), as derived from the power rule for integration: If \( f(y) = y^n \), then its antiderivative is \( \frac{y^{n+1}}{n+1} \), provided \( n eq -1 \).

• An antiderivative essentially 'reverses' differentiation, aiding in evaluating definite integrals.
• Knowing antiderivatives streamlines calculating the area under curves.
• Substituting and converting to a simpler form often helps in identifying the antiderivative more easily.

Solving for the antiderivative is a crucial step in the overall evaluation process of improper integrals.

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals involving square roots or quadratic expressions. In this context, we utilized trigonometric substitution to address the complexity posed by \( \frac{\cos(x)}{\sqrt{\sin(x)}} \).
The key was to transform the variable, replacing \( x \) with a trigonometric function that simplifies the integration process.
For our exercise, by letting \( y = \sin(x) \) and \( dy = \cos(x) \, dx \), we converted the original integral into \( \int_{0}^{1} y^{-1/2} \, dy \), which is far easier to evaluate.
This substitution helped manage the points where the original function might become undefined, essentially eliminating the discontinuity.

• Trigonometric substitution is ideal for integrals with forms involving \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 + a^2} \), or \( \sqrt{x^2 - a^2} \).
• Proper substitution can turn a challenging problem into a solvable one.
• Keep track of changed limits of integration; they're crucial after a substitution.

This approach is particularly useful because trigonometric identities simplify expressions, making integration straightforward.