91Ó°ÊÓ

The Substitution Method is a powerful tool for evaluating integrals, making complex problems more manageable. In this exercise, the goal was to evaluate the iterated integral \( \int_{0}^{\pi / 2} \int_{0}^{1} x \cos(xy) \, dy \, dx \). Using substitution helps simplify the integration process, particularly for the inner integral.

Here’s a brief walkthrough:

• We start by setting a substitution that simplifies the variable expression. In this case, let \( u = xy \), then \( du = x \, dy \). This substitution transforms complex expressions into simpler ones.
• As we substitute \( dy \) with \( \frac{du}{x} \), and adjust the limits of integration for \( y \), the inner integral becomes \( \int_{0}^{x} \cos(u)\frac{du}{x} \).
• The integral \( \frac{1}{x} \int_{0}^{x} \cos(u) \, du \) is now easier to evaluate since it involves a direct integration of the cosine function, a basic trigonometric function.

Substitution simplifies integration by converting functions into more familiar forms, reducing potential errors during calculation.

Integrating by parts is another valuable method for solving integrals, generally used when substitution is impractical. This method wasn't directly applied in our exercise, but it's important to understand how it differs.

The formula is derived from the product rule for differentiation, captured by:\[\int u \, dv = uv - \int v \, du.\]Integration by parts is useful when dealing with products of functions, typically involving polynomials and logarithms or exponentials. Although it wasn’t needed for the cosine function here, it is often necessary when direct substitution doesn't simplify everything.

• Choose parts \( u \) and \( dv \) such that the integration of \( dv \) is simpler.
• Differentiate \( u \) to get \( du \), and integrate \( dv \) to get \( v \).
• Apply the formula, simplifying step by step to solve the integral.

This technique is complementary to substitution, enriching your integral-solving toolkit.

The cosine function, \( \cos(x) \), is a fundamental trigonometric function, appearing frequently in both definite and indefinite integrals. This exercise utilized the cosine function inside the inner integral as part of the iterated process.

The integration rules for cosine are straightforward:

• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The integral of \( \cos(x) \) is \( \sin(x) \) plus a constant for indefinite integrals.

Within the context of definite integrals, the cosine function integrates between specific limits. Here, integrating \( \cos(u) \) over the limits transformed by the substitution yields \( \sin(x) \). This outcome helped facilitate the calculation of the outer integral, which integrated the simple function \( \sin(x) \) over the set boundary from 0 to \( \pi / 2 \), leading to a definitive answer.

Definite integrals provide a concrete numerical result, distinguishing them from indefinite integrals. They are integral in solving real-world problems by evaluating the area under a curve between fixed limits. In the provided iterated integral example, the task was to calculate a definite integral over specified bounds for two variables, \( x \) and \( y \).

Key features of definite integrals:

• You compute with upper and lower limits, which are numerically integrated to find a total area or accumulated quantity.
• The result is a single number rather than a general formula, denoted without an integration constant.
• For multiple integrals, each variable of integration resolves independently within its particular bounds.

In this problem, evaluating the inner integral set the stage for solving the outer definite integral, resulting in the tidy final result of 1. Such problems unravel layers of integration, where understanding definite integrals is crucial.