Problem 18 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(15-22\). $$ \int_{0}^{\pi} \sqrt{1-\cos ^{2} \theta} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Identify the Integrand

We start with the integral \( \int_{0}^{\pi} \sqrt{1 - \cos^2 \theta} \, d\theta \). Notice that the term under the square root, \(1 - \cos^2 \theta\), can be simplified using a trigonometric identity.

Step 2: Simplify the Expression

Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which allows us to rewrite \( 1 - \cos^2 \theta \) as \( \sin^2 \theta \). This simplifies the integrand to \( \sqrt{\sin^2 \theta} \), which is \( |\sin \theta| \).

Step 3: Determine the Absolute Value

Considering the range of integration from \(0\) to \(\pi\), \(\sin \theta\) is non-negative. Therefore, \(|\sin \theta| = \sin \theta\) within this interval.

Step 4: Set up the Simplified Integral

We can now write the integral as \( \int_{0}^{\pi} \sin \theta \, d\theta \).

Step 5: Integrate the Function

The antiderivative of \( \sin \theta \) is \(-\cos \theta\). So, compute \(-\cos \theta\) evaluated from \(0\) to \(\pi\).

Step 6: Evaluate the Definite Integral

Calculate \(-\cos(\pi) + \cos(0) = -(-1) + 1 = 1 + 1 = 2\).

Key Concepts

Trigonometric Identities

When tackling integrals, particularly those involving trigonometric functions, trigonometric identities are immensely valuable. They can transform a complicated expression into something more manageable. Here, our given integral involved the term \( \sqrt{1 - \cos^2 \theta} \). By using the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \), we can simplify this to \( \sin^2 \theta \). This shows how

• Trigonometric identities help reduce complex integrative expressions.
• They provide a bridge to simpler, equivalent forms.

For this problem, revealing that \( 1 - \cos^2 \theta \) simplifies to \( \sin^2 \theta \) allows the square root to cancel with the square, yielding \( |\sin \theta| \), further simplifying the integrative process.

Definite Integrals

Definite integrals calculate the net area under a curve between two specific points on the x-axis. They are essential in various applications, including finding displacement in physics or area under a curve in mathematics. In our exercise:

• We focused on the interval \([0, \pi]\), which represents the bounds of the definite integral.
• The task was to compute the integral of \( \sin \theta \), simplified through our identity, between these bounds.

The evaluation of a definite integral involves taking the antiderivative of the function and then applying the Fundamental Theorem of Calculus. This involves evaluating the antiderivative at the upper bound and subtracting the result at the lower bound. It gives us the total area, taking into account positive and negative sections.

Antiderivatives

Antiderivatives, or indefinite integrals, are functions that describe the accumulation of quantities. Finding an antiderivative is like working backwards from a derivative. For the integral of \( \sin \theta \), its antiderivative is \(-\cos \theta\). This means:

• When \( -\cos \theta \) is differentiated, it returns \( \sin \theta \).
• We use this antiderivative to compute definite integrals.

In our solution, the antiderivative was key to calculating the definite integral. We evaluated \( -\cos \theta \) at the endpoints of the interval from \(0\) to \(\pi\). The process highlights how antiderivatives are the inverse operation of taking derivatives, helping us integrate over specified bounds to find the total accumulation.