Problem 17 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Identify the Integral

We need to evaluate the integral \( \int_{0}^{\pi} \sqrt{1 - \sin^2 t} \, dt \). The expression under the square root, \( 1 - \sin^2 t \), can be recognized as \( \cos^2 t \) due to the Pythagorean identity. This simplifies the integral to \( \int_{0}^{\pi} \sqrt{\cos^2 t} \, dt \).

Step 2: Simplify the Integral

The square root of \( \cos^2 t \) simplifies to \( |\cos t| \), the absolute value of \( \cos t \). This is because square roots return the non-negative root. Since we are integrating from 0 to \( \pi \), \( \cos t \) is non-negative in \( [0, \frac{\pi}{2}] \) and non-positive in \( [\frac{\pi}{2}, \pi] \). Thus, \( |\cos t| \) simplifies to \( \cos t \) in the first half and \( -\cos t \) in the second half.

Step 3: Split the Integral

Split the original integral at \( \frac{\pi}{2} \):\[ \int_{0}^{\pi} |\cos t| \, dt = \int_{0}^{\frac{\pi}{2}} \cos t \, dt + \int_{\frac{\pi}{2}}^{\pi} -\cos t \, dt. \]

Step 4: Evaluate the First Integral

Evaluate \( \int_{0}^{\frac{\pi}{2}} \cos t \, dt \):\[\int_{0}^{\frac{\pi}{2}} \cos t \, dt = [\sin t]_{0}^{\frac{\pi}{2}} = \sin \left( \frac{\pi}{2} \right) - \sin(0) = 1 - 0 = 1.\]

Step 5: Evaluate the Second Integral

Evaluate \( \int_{\frac{\pi}{2}}^{\pi} -\cos t \, dt \):\[\int_{\frac{\pi}{2}}^{\pi} -\cos t \, dt = -[\sin t]_{\frac{\pi}{2}}^{\pi} = -\left( \sin \pi - \sin \left( \frac{\pi}{2} \right) \right) = - (0 - 1) = 1.\]

Step 6: Sum the Results

Add the results of the two integrals:\[1 + 1 = 2.\]Thus, the original integral evaluates to 2.

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry. It states that for any angle \( t \), the relationship \( \sin^2 t + \cos^2 t = 1 \) holds true. This identity arises from the definition of sine and cosine on the unit circle, where the radius is always equal to 1.
When we encounter expressions like \( \sqrt{1 - \sin^2 t} \) in integrals, we can utilize the Pythagorean identity to simplify our work. By rearranging the identity, we see that \( 1 - \sin^2 t \) is equivalent to \( \cos^2 t \).
Understanding and using this identity is crucial when solving integrals involving trigonometric functions because it often allows us to transform complicated expressions into simpler ones.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in calculus for understanding oscillatory motion, wave patterns, and angles. In this exercise, we are focused on the cosine function \( \cos t \), which is

• periodic, repeating its values every \(2\pi\) radians,
• even, meaning \(\cos(-t) = \cos t\),
• bounded, always between -1 and 1.

These properties affect how we handle integrals involving \( \cos t \). When integrating over intervals such as \([0,\pi]\), we must consider these properties and how they impact the integral calculation. In this particular case, understanding that \( \cos t \) changes sign at \( \pi/2 \) is important for correctly handling the absolute value in the transformation of the integral.

Absolute Value

The concept of absolute value is vital when dealing with square roots, especially in trigonometric contexts. Absolute value of a number, say \(x\), is denoted \(|x|\) and is defined as:

• \(+x\) if \(x\) is positive or zero
• \(-x\) if \(x\) is negative

For the function \(|\cos t|\), we need to pay attention to the sign of \(\cos t\) over the interval. In the interval \([0, \pi]\), \(\cos t\) transitions from being positive in \([0, \frac{\pi}{2}]\) to negative in \([\frac{\pi}{2}, \pi]\).
This change requires splitting the integral at \(\frac{\pi}{2}\), allowing us to integrate \(\cos t\) and \(-\cos t\) separately, which simplifies computationally, as seen in the provided solution.

Calculus Techniques

Calculus techniques are essential tools for solving integrals, especially when we deal with functions that involve trigonometric expressions. Breaking down integrals is a common technique when dealing with piecewise functions. For instance, the integral \( \int_{0}^{\pi} |\cos t| \, dt \) was split based on the behavior of \( \cos t \) over the interval \([0,\pi]\).
Other techniques such as substitution or recognizing symmetry can often simplify the processes. Here, simplifying \( \sqrt{\cos^2 t} \) to \( |\cos t| \) is a form of function transformation using the properties of square roots and trigonometric identities. Utilizing these techniques effectively requires understanding how integrals work and how trigonometric functions behave, ensuring the solution is precise and accurate.