Question

Evaluate the given integral. $$ \int_{-\pi / 2}^{\pi / 2} \cos ^{2} x \sin x d x $$

Short answer

The integral evaluates to 0.

Step-by-step solution

Identify the Integral Components

The integral we need to solve is \( \int_{-\pi/2}^{\pi/2} \cos^2 x \sin x \, dx \). Here, our function consists of \( \cos^2 x \) and \( \sin x \) which indicates a trigonometric integral.

Choose a Substitution Method

For this integral, notice that the derivative of \( \cos x \) is \( -\sin x \). This suggests that a substitution might be beneficial. Let's set \( u = \cos x \), then \( du = -\sin x \, dx \). This makes the integral simpler to solve.

Change the Bounds and Substitute Variables

Change the bounds according to substitution: When \( x = -\pi/2 \), \( u = \cos(-\pi/2) = 0 \). When \( x = \pi/2 \), \( u = \cos(\pi/2) = 0 \). Substitute into the integral: \[ -\int_{0}^{0} u^2 \, du \].

Evaluate the Integral

The integral from \( 0 \) to \( 0 \) of any function is zero due to the zero width of the interval. So, the integral \(-\int_{0}^{0} u^2 \, du = 0 \).

Verify and Conclude

Double-check each transformation and calculation to ensure the substitution was done correctly. Conclude that the result of the original integral is zero.

Key concepts

Substitution Method

The substitution method is a handy tool when dealing with integrals, especially those involving trigonometric functions like cosine and sine. In the given problem, we need to evaluate the integral \( \int_{-\pi / 2}^{\pi / 2} \cos^2 x \sin x \, dx \).
To simplify this integral, the idea is to substitute a part of the integral to transform it into an easier form. Here, notice how the derivative of \( \cos x \) is \( -\sin x \). This minor detail is a hint that swapping \( \cos x \) for a new variable, say \( u \), can simplify our work.

Here's how it works: we set \( u = \cos x \), which automatically gives \( du = -\sin x \, dx \). With this substitution, the integral becomes easier to handle because \( \cos x \) and \( \sin x \) pair up perfectly for substitution.

Integral Bounds Transformation

When we substitute a new variable into our integral, it changes not just the variable inside the integral but also the boundaries. Transforming the integral bounds is crucial to maintain the accuracy of our integration.

In our example, originally the integral has bounds from \( x = -\pi/2 \) to \( x = \pi/2 \). But when we substitute \( u = \cos x \), we need to revise these limits according to our new variable \( u \).

• When \( x = -\pi/2 \), \( u = \cos(-\pi/2) = 0 \)
• When \( x = \pi/2 \), \( u = \cos(\pi/2) = 0 \)

Thus, the integral's new bounds are from \( u = 0 \) to \( u = 0 \). It's important to note that if both lower and upper bounds turn out to be the same, the integral will evaluate to zero.

Integral Evaluation

The final step in solving any integral is evaluating it, which is much easier once the integral has been simplified. In our problem, after substituting and transforming the bounds, we ended up with:\[ -\int_{0}^{0} u^2 \, du \]

Even though the integrand \( u^2 \) isn't complex, actually evaluating this integral is simple since the interval width is zero.

• An integral over an interval with zero width like from \( u = 0 \) to \( u = 0 \) will always result in zero. This holds for any function, not just trigonometric ones.

Thus, the solved integral confirms that the value is zero, perfectly wrapping up our problem without additional computations.