Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals. $$\int_{-2 \pi}^{2 \pi} \cos \frac{\theta}{8} d \theta$$

Short answer

Question: Evaluate the definite integral $$\int_{-2 \pi}^{2 \pi} \cos \frac{\theta}{8} d \theta.$$ Answer: The definite integral is equal to $$8\sqrt{2}.$$

Step-by-step solution

Identify the Change of Variable, Express dθ

We are given the following integral with a change of variable: $$\int_{-2 \pi}^{2 \pi} \cos \frac{\theta}{8} d \theta$$ Now we need to identify the variable change. Here, the change of variable can be represented as $$u = \frac{\theta}{8}.$$ We differentiate u with respect to θ to solve for dθ: $$d \theta = 8 du$$

Substituting the Change of Variable and Changing the Limits of Integration

We will now substitute the value of u and the value of dθ we just found into the original integral: $$\int_{-2 \pi}^{2 \pi} \cos \frac{\theta}{8} d \theta$$ Substitute $$u = \frac{\theta}{8}$$ and $$d \theta = 8 du$$ to get: $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos u \cdot 8du$$ Notice that we change the limits of integration from -2π and 2π to -π/4 and π/4, respectively. This is because when we substitute u, we must also change our limits of integration accordingly: $$u = \frac{-2 \pi}{8} = -\frac{\pi}{4}$$ $$u = \frac{2 \pi}{8} = \frac{\pi}{4}$$

Evaluate the Integral

Now we have a simplified integral and can evaluate it as follows: $$8\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos u du$$ This integral can be easily evaluated as we know the antiderivative of the cosine function, which is the sine function. Thus, we evaluate it as: $$8[\sin u]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = 8 (\sin(\frac{\pi}{4}) - \sin(-\frac{\pi}{4})) = 8 (2 \sin(\frac{\pi}{4}))$$

Calculate the Final Result

Using the sine value, we compute the final result: $$8 (2 \sin(\frac{\pi}{4})) = 8 (2 \cdot \frac{\sqrt{2}}{2}) = 8\sqrt{2}$$ Thus, the definite integral of $$ \int_{-2 \pi}^{2 \pi} \cos \frac{\theta}{8} d \theta $$ is equal to $$8\sqrt{2}.$$

Key concepts

Change of Variables

In calculus, the change of variables technique is a powerful tool used to simplify the process of evaluating integrals. The goal is to transform a difficult integral into a more manageable form. In our given exercise, we have the integral \[ \int_{-2 \pi}^{2 \pi} \cos \frac{\theta}{8} \, d\theta \] By introducing a change of variables, we redefine the expression in terms of a new variable, \( u \). In this case, we set \( u = \frac{\theta}{8} \) to simplify the expression inside the cosine function. After establishing our new variable, we find the differential \( du \). By differentiating \( u \) with respect to \( \theta \), we find \( du = \frac{1}{8} \, d\theta \), which rearranges to \( d\theta = 8 \, du \). This substitution allows us to rewrite the original integral in terms of \( u \), making it easier to solve. Remember, changing variables effectively shifts the problem into a different perspective, helping us unlock more complicated integrals with ease. The substitution is not only used to modify the variable itself but also affects the differential \( d\theta \) and the limits of integration, transitioning the whole problem to a new variable landscape.

Limits of Integration

When performing a change of variables in a definite integral, it is crucial to adjust the limits of integration to correspond with the new variable. In our exercise, the original limits of integration are \( -2 \pi \) and \( 2 \pi \). These need to be properly converted to match our substituted variable \( u \) that we established as \( u = \frac{\theta}{8} \).This adjustment is straightforward. For the lower limit, where \( \theta = -2\pi \), we substitute into \( u = \frac{\theta}{8} \) resulting in \( u = -\frac{\pi}{4} \). The same process applies for the upper limit of \( \theta = 2\pi \), which converts to \( u = \frac{\pi}{4} \).By changing the limits of integration, the entire integral stays consistent with the new variable and accurately reflects the original problem's "domain." Neglecting this step would lead to incorrect results, as the integral's bounds would misalign with the terms being integrated.

Cosine Function

The cosine function is a fundamental trigonometric function that appears often in calculus, especially in integrations. In this exercise, we are integrating the cosine function, specifically \( \cos \frac{\theta}{8} \). The properties of the cosine function make it periodic and symmetric, which can greatly simplify integration over certain intervals.Cosine is periodic with a period of \( 2\pi \), meaning that every \( 2\pi \), the function repeats its values. Additionally, cosine is an even function, thus satisfying \( \cos(-x) = \cos(x) \). These properties ensure that integrating the cosine function results in predictable and often simple results.Moreover, when the integral of the cosine function is evaluated over symmetric limits (such as \(-\pi/4\) to \(\pi/4\)), it helps in directly applying its antiderivative. Recognizing these characteristics assists in simplifying the definite integrals and accurately solving the problem with the correct application of integral properties.

Antiderivative

The antiderivative is a key concept in solving integration problems. It is essentially the reverse process of differentiation. For any function \( f(x) \), an antiderivative is a function \( F(x) \) such that the derivative of \( F(x) \) is \( f(x) \). In our integral, \[ 8\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos u \, du, \] we find that the antiderivative of \( \cos u \) is \( \sin u \). Once we establish this, we evaluate it from \( -\pi/4 \) to \( \pi/4 \). What makes antiderivatives so powerful is their ability to transform complex integral evaluations into simple arithmetic operations using the Fundamental Theorem of Calculus. This theorem states that the definite integral of a function over an interval \([a, b]\) can be found using its antiderivative:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \].Thus, in this exercise, we compute the final outcome by calculating \( 8 [\sin(\frac{\pi}{4}) - \sin(-\frac{\pi}{4})] \), leading to the final result \( 8 \sqrt{2} \). The antiderivative provides a straightforward path from the derivative back to the original function, making them vital in calculus.