Problem 72 - Calculus: Early Transcendentals

Question

Evaluate the definite integral. \( \displaystyle \int^{T/2}_0 \sin (2\pi t/T - \alpha) \, dt \)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Change of Variables

To evaluate the integral, let's start by making a substitution to simplify the integral. We can let \( u = \frac{2\pi t}{T} - \alpha \). This gives us \( du = \frac{2\pi}{T} dt \) or \( dt = \frac{T}{2\pi} du \). We also need to change the limits of integration. When \( t = 0 \), \( u = -\alpha \) and when \( t = \frac{T}{2} \), \( u = \pi - \alpha \). Thus, the integral becomes: \[ \int_{-abla}^{\pi - \alpha} \sin(u) \frac{T}{2\pi} \, du \]

Step 2: Solve the Modified Integral

The integral \( \int \sin(u) \, du \) is straightforward to solve. Its antiderivative is \(-\cos(u)\). Therefore, integrating \( \sin(u) \frac{T}{2\pi} \) gives: \[-\frac{T}{2\pi} \cos(u)\]

Step 3: Evaluate the Definite Integral

Now, evaluate this expression at the new limits of integration. Substituting the limits \(-\alpha\) and \(\pi - \alpha\), we have: \[ -\frac{T}{2\pi} [\cos(\pi - \alpha) - \cos(-\alpha)] \]Using the identity \( \cos(\pi - \alpha) = -\cos(\alpha) \), it simplifies to: \[ -\frac{T}{2\pi} [-\cos(\alpha) + \cos(\alpha)] = 0 \]

Step 4: Conclusion: Final Calculation

After simplifying the expression, we find that the integral evaluates to zero because the two terms in the expression cancel each other out.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in integral calculus that calculates the net area under a curve over a specific interval. For example, when evaluating the integral \( \int_a^b f(x) \, dx \), you determine the total area between the function \( f(x) \) and the x-axis from \( x = a \) to \( x = b \). This process transforms functions into numerical outcomes that represent this area.

Key aspects to understand about definite integrals include:

• Limits of Integration: The numbers \( a \) and \( b \) are called the 'lower' and 'upper' limits, respectively.
• Net Area: The integral evaluates the net area; thus, it can be positive, negative, or zero, depending on how the areas above and below the x-axis behave.

In the context of our problem, we're specifically integrating \( \sin(2\pi t/T - \alpha) \) from \( 0 \) to \( T/2 \), calculating the net sinusoidal area for half of its period.

Change of Variables

The 'Change of Variables' method, also known as substitution, is a powerful tool in integration aimed at simplifying complex integrals into more manageable forms. This often involves substituting a part of the integrand and replacing the differential accordingly.

For example, in our problem, we substitute \( u = \frac{2\pi t}{T} - \alpha \) to simplify the sine function within the integral. This substitution changes the variables and limits:

• Substitute: Replace \( t \) with the new variable \( u \)
• Adjust the differential: \( dt = \frac{T}{2\pi} du \), ensuring the integral remains balanced.
• Modify limits: Calculate new limits from the original \( t \) range.

This results in a transformed integral \( \int_{-\alpha}^{\pi - \alpha} \sin(u) \frac{T}{2\pi} \, du \) that is often simpler to evaluate.

Antiderivative

Finding an antiderivative, also called an indefinite integral, is about determining a function whose derivative is the original function given. In a definite integral, evaluating an antiderivative allows us to calculate the total accumulated change over an interval.

For trigonometric functions like \( \sin(u) \), the antiderivative is straightforward:

• \( \int \sin(u) \, du = -\cos(u) + C \) where \( C \) represents a constant of integration.

In definite integrals, we compute the antiderivative and then substitute the limits to find the total area. In our case, after substitution, this process leads to:

• Antiderivative: \( -\frac{T}{2\pi} \cos(u) \)
• Evaluate at limits \( -\alpha \) and \( \pi - \alpha \)

These steps convert the definite integral into a manageable computation, highlighting how essential antiderivatives are to this process.

Trigonometric Integration

Trigonometric integration involves integrating functions with trigonometric components, such as sine and cosine. Understanding these functions' properties is crucial as it allows for intuitive problem-solving.

Consider some essential insights for trigonometric integration:

• Basic Antiderivatives: Knowing that \( \int \sin(x) \, dx = -\cos(x) \) and \( \int \cos(x) \, dx = \sin(x) \) simplifies your calculus problems.
• Use of identities: Trigonometric identities can simplify complex expressions during integration. For instance, during our calculation, \( \cos(\pi - \alpha) = -\cos(\alpha) \) helped simplify the integral's outcome.
• Recognize Symmetry: Symmetry in integrals with limits can sometimes lead to cancellations, like our example where everything simplifies to zero.

Grasping these essentials can aid in effectively tackling trigonometric integrals, making complex sinusoidal functions manageable.