Question

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. \(\int_{-\pi}^{\pi} \cos ^{2}(3 x) d x\)

Short answer

The definite integral evaluates to \( \pi \).

Step-by-step solution

Use a Trigonometric Identity

To simplify the integrand \( \cos^2(3x) \), we'll use the trigonometric identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \). Applying this to \( \cos^2(3x) \), we get: \[ \cos^2(3x) = \frac{1 + \cos(6x)}{2} \]. Therefore, the integral becomes: \[ \int_{-\pi}^{\pi} \frac{1 + \cos(6x)}{2} \, dx \].

Split the Integral

Split the integral \( \int_{-\pi}^{\pi} \frac{1 + \cos(6x)}{2} \, dx \) into two separate integrals: \[ \frac{1}{2} \left( \int_{-\pi}^{\pi} 1 \, dx + \int_{-\pi}^{\pi} \cos(6x) \, dx \right) \]. This will make the computation simpler.

Evaluate the First Integral

The first integral is \( \int_{-\pi}^{\pi} 1 \, dx \), which evaluates to the length of the interval. Therefore, the result is \( 2\pi \).

Evaluate the Second Integral

The second integral \( \int_{-\pi}^{\pi} \cos(6x) \, dx \) is zero, because \( \cos(6x) \) is an even function and the limits of integration are symmetric around zero. The integral of an even function over such an interval is symmetric and has equal areas above and below the x-axis, which results in zero.

Combine the Results

Substitute the results of the two integrals back: \[ \frac{1}{2} \left( 2\pi + 0 \right) = \frac{1}{2} \times 2\pi = \pi \]. Therefore, the value of the definite integral is \( \pi \).

Key concepts

Trigonometric Identity

A trigonometric identity helps us simplify complex expressions involving trigonometric functions. In integral calculus, these identities are often used to simplify the integrand, which is the function being integrated. One particularly useful identity is the double-angle formula for cosine, expressed as: \[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \] This identity allows us to rewrite \( \cos^2(3x) \) as \( \frac{1 + \cos(6x)}{2} \). This expression is much easier to integrate, especially within definite integrals. Using identities like this can transform a complicated integral into something more manageable, which is key to solving many problems in calculus. Remember, recognizing and applying the right trigonometric identity is an essential skill in dealing with integrals effectively.

Even Function

An even function is a function where the graph is symmetric with respect to the y-axis. This means that for an even function, \( f(x) = f(-x) \) for all values of \( x \). Examples include \( \cos(x) \) and \( x^2 \). When evaluating definite integrals involving even functions, it's useful to consider the properties of symmetry. For instance, the integral of \( \cos(6x) \) from \( -\pi \) to \( \pi \) is zero because the positive and negative areas cancel each other out due to symmetry. Essentially, every segment of area beneath the x-axis has its counterpart of equal magnitude above the x-axis, resulting in a net area of zero. This property greatly simplifies computations, especially over symmetric intervals.

Symmetric Interval

In integral calculus, a symmetric interval occurs when the limits of integration are equal in magnitude but opposite in sign, like \( -\pi \) to \( \pi \). This symmetry often allows for simplifications, particularly when dealing with even or odd functions. For even functions, the symmetric properties mean that the area from the negative to positive side is canceled out, leading to potentially significant simplifications. When you see a symmetric interval, it can be a cue to look for even or odd functions in the integrand. This can help to quickly evaluate the integral without detailed calculations. In problems where symmetry plays a crucial role, spotting these intervals can save you a lot of time and effort, allowing you to focus on more complex aspects of the problem.

Integral Calculus

Integral calculus is a major branch of calculus focused on accumulation and the calculation of areas under curves. Definite integrals, such as the one described in this exercise, calculate the net area within specified bounds. They can be used to determine many physical properties, such as displacement, area, volume, and total consumption. The process involves finding the antiderivative (or integral) of a function and then applying the limits of integration to find the total area. Knowledge of trigonometric identities, properties of even and odd functions, and symmetric intervals are incredibly useful in simplifying integrals and obtaining precise results. By practicing integral calculus, you become adept at not only solving specific problems but also gaining deeper insights into the behavior of various mathematical functions. This integration process blends these theoretical concepts with practical applications, making it a powerful tool in math and science.