Problem 53 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{-\pi}^{\pi} \cos ^{2} x d x$$

Step-by-Step Solution

Answer

Answer: The value of the integral is \(\pi\).

Step 1: Use the double-angle identity for cosine

Rewrite the cosine squared term using the double-angle identity for cosine: $$\cos 2x = 2 \cos^2 x - 1$$ Then, solve for \(\cos^2 x\): $$\cos^2 x = \frac{1 + \cos(2x)}{2}$$ The given integral now becomes: $$\int_{-\pi}^{\pi} \frac{1+\cos(2x)}{2} dx$$

Step 2: Find the antiderivative

Split the integral into two separate integrals: $$\frac{1}{2}\int_{-\pi}^{\pi} 1 dx + \frac{1}{2}\int_{-\pi}^{\pi} \cos(2x) dx$$ Find the antiderivative of the first integral: $$\frac{1}{2}\int_{-\pi}^{\pi} 1 dx = \frac{1}{2}[x]_{-\pi}^{\pi}$$ Find the antiderivative of the second integral: $$\frac{1}{2}\int_{-\pi}^{\pi} \cos(2x) dx = \frac{1}{4}\int_{-\pi}^{\pi} \frac{d(\sin(2x))}{dx} dx = \frac{1}{4}[\sin(2x)]_{-\pi}^{\pi}$$

Step 3: Apply the limits of integration

Evaluate the antiderivatives at their respective bounds. For the first integral: $$\frac{1}{2}[x]_{-\pi}^{\pi} = \frac{1}{2}(\pi - (-\pi)) = \pi$$ For the second integral: $$\frac{1}{4}[\sin(2x)]_{-\pi}^{\pi} = \frac{1}{4}(\sin(2\pi) - \sin(-2\pi)) = \frac{1}{4}(0-0)=0$$

Step 4: Sum the results and find the final answer

Add the results of the two integrals: $$\pi + 0 = \pi$$ Thus, the value of the integral is: $$\int_{-\pi}^{\pi} \cos ^{2} x dx = \pi$$

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial when it comes to solving integrals that involve trigonometric functions. These identities are transformations that allow us to write trigonometric expressions in different ways. One important identity used for this exercise is the double-angle identity for cosine:

• \(\cos 2x = 2 \cos^2 x - 1\)

This identity can be rearranged to solve for \(\cos^2 x\):

• \(\cos^2 x = \frac{1 + \cos(2x)}{2}\)

This transformation simplifies the process of integration by changing a power of cosine function into a more manageable expression. Simplifying integrals using identities is a key technique in calculus, enabling us to explore solutions that might otherwise seem complex or intractable. Knowing and applying these identities helps to reduce the algebraic complexity, making integration more straightforward.

Antiderivatives

An antiderivative is the reverse process of differentiation. In simple terms, it helps us to find the original function from its derivative. In integration, solving an antiderivative means getting back to the function that was originally differentiated to give the function we are now integrating. In calculus, you'll often hear an antiderivative referred to as the indefinite integral. Let's break down the integral:

• \( \frac{1}{2}\int_{-\pi}^{\pi} 1 \, dx \)

The antiderivative of 1 with respect to \(x\) is \(x\), hence the first part of the integral simply evaluates to \(\frac{1}{2}[x]_{-\pi}^{\pi}\).

• \( \frac{1}{2}\int_{-\pi}^{\pi} \cos(2x) \, dx \)

For the cosine term, notice how the integral \(\int \cos(2x) \, dx\) leads to \(\frac{1}{2} \sin(2x)\). Here, we've substituted \(\cos(2x)\) with \(\sin(2x)\) divided by its derivative (\(2\)). This is often a necessary technique when finding antiderivatives involving trigonometric functions, translating a derivative into its primitive form.

Integration Limits

Integration limits define the range over which the function is being integrated. They effectively "bound" the integral, indicating the start and end points for the area under the curve. These limits are crucial because they directly affect the integral's value, especially in definite integrals. In this exercise, our integration limits are \(-\pi\) to \(\pi\). Once we find the antiderivatives, as shown previously, we apply these limits:

• \( \frac{1}{2}[x]_{-\pi}^{\pi} = \frac{1}{2}(\pi - (-\pi)) = \pi \)
• \( \frac{1}{4}[\sin(2x)]_{-\pi}^{\pi} = \frac{1}{4}(\sin(2\pi) - \sin(-2\pi)) = 0 \)

Apply integration limits by substituting them into the antiderivative and calculating the difference or sum, as suitable. In this case, the limits translate to straightforward evaluation as \(\pi\) and \(-\pi\) effectively cover a symmetric interval around zero, especially useful for trigonometric functions with periodic nature like \(\sin\) and \(\cos\). Remember that knowing how to apply these boundaries is as fundamental as finding the integral itself.