91Ó°ÊÓ

Understanding the **region of integration** is the first important step in solving a double integral problem. The region is defined by boundaries that outline where the integration takes place. In this exercise, the region is bounded by two curves:

• The upper boundary: \( y = \sin x \).
• The lower boundary: \( y = 0 \) (which is the x-axis).

These boundaries confine the area under the \( \sin x \) curve between \( x = 0 \) and \( x = \pi \). This means you integrate over a "slice" of the x-axis between these two points. Essentially, you move from left to right, considering how y varies from 0 to \( \sin x \) at each point x. Understanding this setup is crucial, as it determines the limits of integration for the double integral, allowing you to write the problem correctly as an iterated integral.

Integration by parts is a technique often used to solve integrals that are products of functions. The formula comes from the product rule for differentiation and is given for integral: \[\int u \, dv = uv - \int v \, du.\] In this exercise, we apply integration by parts to evaluate \( \int x\sin x \, dx \). Here's a step-by-step way to select and use the formula:

• Choose \( u = x \) because it simplifies when differentiated, and \( dv = \sin x \, dx \) because its integral is simple.
• The differentiation of \( u \) gives \( du = dx \) and the integration of \( dv \) yields \( v = -\cos x \).
• Apply the integration by parts formula to get \( -x\cos x + \int \cos x \, dx \), leading to \( -x\cos x + \sin x \).

Evaluate this result from 0 to \( \pi \), giving you the contribution of this part of the integral. Learning how to choose \( u \) and \( dv \) properly comes with practice and is a valuable technique for tackling more complex integrals.

When integrating expressions involving trigonometric functions, like \( \sin^2 x \), leveraging **trigonometric identities** can simplify the process. One commonly used identity is: \[\sin^2 x = \frac{1 - \cos 2x}{2}.\] This identity allows you to transform \( \sin^2 x \) into a form that is easier to integrate. The integral \( \int \frac{\sin^2 x}{2} \, dx \) can thus be rewritten using this identity as: \[\int \frac{1}{4} (1 - \cos 2x) \, dx.\] This expression simplifies into two separate integrals \( \int \frac{1}{4} \, dx \) and \( \int -\frac{1}{8}\cos 2x \, dx \). These are more straightforward to evaluate. Such trigonometric identities are incredibly useful for solving a broad range of integration problems, not just limited to \( \sin^2 x \). Understanding and applying these identities can greatly simplify your work with trigonometric integrals.