91Ó°ÊÓ

When calculating integrals, recognizing symmetry can simplify the process significantly. Symmetry often determines whether the integral will have an easy solution. In the exercise, you're dealing with the function \(\cos^2 \theta \sin \theta\) over a specific symmetric interval: \([-\pi, \pi]\).

Always check if a function is even or odd about an interval symmetric around zero:

• **Even Functions:** These satisfy \(f(-x) = f(x)\) and typically result in non-zero integrals over symmetrical intervals.
• **Odd Functions:** These have the property \(f(-x) = -f(x)\). Their integrals over symmetric intervals around zero equal zero.

Applying this to the present challenge, because \(\cos^2 \theta \sin \theta\) is odd, the evaluated integral from \(-\pi\) to \(\pi\) is zero.

This is because all the positive areas generated by the function on one side of the y-axis are canceled by the negative areas on the opposite side. By spotting this symmetry, you save time and effort in calculations.

The substitution method is a popular technique used to simplify certain integrals, making them easier to evaluate. It's much like changing variables in an equation to make it more straightforward. In our example, the substitution method turns the task into an easier problem by transforming the variable \(\theta\) into a new variable, \(u\).

Here's a basic breakdown of how substitution worked in the exercise:

• **Start with Function:** \(\cos^2 \theta \sin \theta\).
• **Choose a Substitution:** Let \(u = \cos \theta\), then the derivative \(du = -\sin \theta \, d\theta\).
• **Change the Limits:** As \(\theta\) transforms, so too must the integral's limits: \(u = 1\) for \(\theta = 0\) and \(u = -1\) for \(\theta = \pi\).
• **Transform the Integral:** The original integral changes into \(-\int_{1}^{-1} u^2 \, du\). Flip the limits to resulting in \(\int_{-1}^{1} u^2 \, du\).

This substitution simplifies the problem significantly, allowing you to evaluate the integral of \(u^2\) with ease, yielding the result \(\frac{2}{3}\).

Substitution not only simplifies complex expressions but can also change the approach to solving the integral altogether.

Finding the area under a curve is a fundamental application of definite integrals. It calculates the space between the curve and the x-axis over a defined interval. In terms of physics, it can represent total distance, work done, etc., depending on what the curve represents.

For part (b) of the exercise, you're calculating this area for the function \(y=\cos^2 \theta \sin \theta\) over \(\theta = 0\) to \(\theta = \pi\). As illustrated, steps 3 and 4 of the solution simplify the process using substitution, resulting in an area value of \(\frac{2}{3}\) square units.

Understanding the concept in-depth involves realizing that:

• **Area Calculation:** Integral evaluation directly translates to area by summing infinitesimally small slices under the curve.
• **Positive and Negative Areas:** Be aware, especially with periodic functions like sine and cosine, that areas above the x-axis are positive while those below are negative.

Overall, determining areas helps in visualizing the results of an integral, and integrating familiarizes us with different techniques to approach diverse mathematical problems.