Question

Evaluate the integral and subsequent iterated integral. (a) \(\int_{0}^{y}(\cos x \sin y) d x\) (b) \(\int_{0}^{\pi} \int_{0}^{y}(\cos x \sin y) d x d y\)

Short answer

(a) \( \sin^2 y \); (b) \( \frac{\pi}{2} \).

Step-by-step solution

Set Up the Integral for Part (a)

The integral in question is \( \int_{0}^{y} \cos x \sin y \, dx \). We treat \( \sin y \) as a constant because we are integrating with respect to \( x \).

Integrate with Respect to x

To find \( \int_{0}^{y} \cos x \sin y \, dx \), we find the antiderivative of \( \cos x \sin y \). This is \( \sin y \int \cos x \, dx = \sin y \cdot \sin x + C \). Thus, \( \int_{0}^{y} \cos x \sin y \, dx = \sin y \sin x \bigg|_{0}^{y} \).

Evaluate the Limits for Part (a)

Evaluate the result from Step 2 from 0 to \( y \). \( \sin y \sin x \bigg|_{0}^{y} = \sin y (\sin y - \sin 0) = \sin^2 y - 0 = \sin^2 y \).

Set Up the Iterated Integral for Part (b)

For part (b), the problem is \( \int_{0}^{\pi} \left( \int_{0}^{y} \cos x \sin y \, dx \right) dy \). Replace the inner integral with the result of part (a): \( \int_{0}^{\pi} \sin^2 y \, dy \).

Use Trigonometric Identity

Use the identity \( \sin^2 y = \frac{1 - \cos(2y)}{2} \) to simplify the integral: \( \int_{0}^{\pi} \frac{1 - \cos(2y)}{2} \, dy \).

Split the Integral

Split the integral as follows: \( \frac{1}{2} \int_{0}^{\pi} (1 - \cos(2y)) \, dy = \frac{1}{2} \left( \int_{0}^{\pi} 1 \, dy - \int_{0}^{\pi} \cos(2y) \, dy \right) \).

Evaluate Each Integral

Compute \( \int_{0}^{\pi} 1 \, dy = [y]_{0}^{\pi} = \pi \). For the second integral, \( \int_{0}^{\pi} \cos(2y) \, dy = \left[ \frac{1}{2} \sin(2y) \right]_{0}^{\pi} = 0 \) since sine of multiples of \( \pi \) is zero.

Combine Results

Combine the results to find: \( \frac{1}{2} \left( \pi - 0 \right) = \frac{\pi}{2} \).

Key concepts

Trigonometric identities

Trigonometric identities play a crucial role in simplifying expressions and solving integrals, especially when dealing with functions that include sine and cosine. These identities convert complex trigonometric expressions into simpler forms. For instance, the identity \( \sin^2 y = \frac{1 - \cos(2y)}{2} \) is immensely useful. It allows us to transform the square of sine into an expression that's easier to integrate. In this integral problem, using such an identity simplifies the process greatly. Remember:

• These identities are fundamental tools in calculus.
• Their primary utility is in integration and differentiation of trigonometric functions.

Mastering these identities can help make seemingly tough integral problems much more manageable. Applying them the right way makes finding solutions less time-consuming and more straightforward.

Antiderivatives

Antiderivatives, or indefinite integrals, are essential in calculus, as they represent the reverse process of differentiation. In this task, finding the antiderivative of \( \cos x \) was necessary because it allowed us to evaluate the inner integral when dealing with iterated integrals.
The key to understanding antiderivatives is recognizing that for a given function, its antiderivative is another function whose derivative is the original function.

• The antiderivative of \( \cos x \) is \( \sin x + C \).
• This result helps to compute definite integrals by evaluating the antiderivative over a specific interval.

Recall that when calculating definite integrals, constants of integration \( C \) cancel out. This leads to simplification when applying limits. The practice with antiderivatives strengthens understanding of fundamental calculus concepts.

Definite integrals

Definite integrals provide the area under a curve within specified limits and are a staple of calculus applications. They are different from indefinite integrals in that they have boundaries set on the variable of integration, which is key in determining a specific numerical value.
In this exercise, the integral \( \int_{0}^{y} \cos x \sin y \, dx \) was evaluated as a definite integral from 0 to \( y \). By doing this, we convert our problem into precisely finding the numerical area using these set intervals.

• Definite integrals often involve using antiderivatives to evaluate.
• The Fundamental Theorem of Calculus connects differentiation and integration, enabling the evaluation of these integrals.

This exercise, specifically calculating \( \int_{0}^{\pi} \), illustrates how definite integrals can help solve multistep integral problems effectively.

Multivariable calculus

Multivariable calculus extends calculus concepts to functions of more than one variable. It's useful in diverse fields, including physics and engineering. Iterated integrals are a cornerstone of multivariable calculus, allowing you to integrate functions with respect to multiple variables sequentially.
In this exercise, we executed iterated integrals by starting with the inner integral and then moving to the outer, \( \int_{0}^{\pi} \int_{0}^{y} \cos x \sin y \, dx \, dy \). This demonstrates:

• The process of solving each integral one at a time in a specific order.
• Simplification at each step through substitutions and evaluated bounds.

Understanding multivariable calculus helps in comprehending more complex systems. It provides tools for working with real-world problems where variables interact intricately.