Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int_{-\pi}^{\pi}(\sin x+\cos x)^{2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(2\pi\).

Step 1: Expand the Expression

We start by expanding the integrand \((\sin x + \cos x)^2\). Using the formula \((a+b)^2 = a^2 + 2ab + b^2\), we have: \[(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x.\]

Step 2: Simplify Using Trigonometric Identities

Recall the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). Substitute this into our expression: \[(\sin^2 x + \cos^2 x) + 2\sin x \cos x = 1 + 2\sin x \cos x.\] This simplifies our integrand to \(1 + 2\sin x \cos x\).

Step 3: Simplify Further Using Product-to-Sum Identity

Use the identity for products of sines and cosines: \(2\sin x \cos x = \sin 2x\). This changes our integrand to \[1 + \sin 2x.\]

Step 4: Integrate Term by Term

We now integrate each term separately from \(-\pi\) to \(\pi\):\[\int_{-\pi}^{\pi} (1 + \sin 2x) \, dx = \int_{-\pi}^{\pi} 1 \, dx + \int_{-\pi}^{\pi} \sin 2x \, dx.\]The integral of \(1\) over \([-\pi, \pi]\) is:\[\int_{-\pi}^{\pi} 1 \, dx = [x]_{-\pi}^{\pi} = \pi - (-\pi) = 2\pi.\]For \(\int_{-\pi}^{\pi} \sin 2x \, dx\), since \(\sin 2x\) is an odd function, the integral over a symmetric interval about zero is zero:\[\int_{-\pi}^{\pi} \sin 2x \, dx = 0.\]

Step 5: Combine the Results

Add the results from the integrated terms to find the total integral:\[\int_{-\pi}^{\pi} (\sin x + \cos x)^2 \, dx = 2\pi + 0 = 2\pi.\]

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools for simplifying expressions that involve sine and cosine functions. They allow us to transform complicated trigonometric equations into more manageable forms. In our exercise, we used the Pythagorean identity, a fundamental formula given by \(\sin^2 x + \cos^2 x = 1\). This identity is a cornerstone in trigonometry, stating that the sum of the squares of sine and cosine functions equals one. It is particularly helpful when dealing with integrals, as it simplifies our task by reducing the complexity of the expression involved.

• Why It's Important: Trigonometric identities not only simplify calculations but also reveal the intrinsic properties of trigonometric functions, making them invaluable for integrals, derivatives, and solving trigonometric equations.

Once this identity is substituted into more complex expressions, it reduces the need for further complicated manipulation, leading to quicker and more straightforward solutions.

Product-to-Sum Formula

The product-to-sum formula is another advantageous tool in trigonometry. It transforms products of sine and cosine functions into simpler sums. In the discussed exercise, the identity \(2\sin x \cos x = \sin 2x\) was employed. This equation turns the product of \(\sin x\) and \(\cos x\) into a single sine term.

• Simplification: The transformation helps in simplifying the integrands which are in multiplied forms, making the integration process more straightforward.
• Usefulness: Such transformations are particularly useful when dealing with integrals over a period, where periodic functions tend to cancel or simplify naturally.

In our exercise, converting \(2\sin x \cos x\) to \(\sin 2x\) allowed us to seamlessly integrate over the desired interval, emphasizing the practicality of the product-to-sum formula in mathematical analysis.

Even and Odd Functions

When evaluating definite integrals, the concepts of even and odd functions play a crucial role, particularly over symmetric intervals.

• Definitions: Even functions have the property that \(f(-x) = f(x)\) and are symmetric about the y-axis. Odd functions satisfy \(f(-x) = -f(x)\) and are symmetric about the origin.

In our solution, \(\sin 2x\) is identified as an odd function. When integrating odd functions over an interval that is symmetric about zero, like \([-\pi, \pi]\), the integral sums to zero. This property drastically simplifies calculations, as evident in the given solution.

• Simplification for Odd Functions: Recognizing that the integral of an odd function over such an interval results in zero can speed up the integration process significantly.

Definite Integrals

Definite integrals calculate the net area under a curve within a specified interval. They are a fundamental concept in calculus, used to determine accumulated quantities, such as area, volume, and in this context, trigonometric functions over a range.

• Evaluation: To compute a definite integral, it involves calculating the antiderivative of the function, followed by applying the limits of integration.
• Application: In our problem, we integrated the function \(1 + \sin 2x\) from \([-\pi, \pi]\), evaluating the simple constant function and applying symmetry for the sine term.

The integration process effectively boils down the problem to computing the change over the specified domain, as seen with the net result of \(2\pi\) for the problem. Understanding how to evaluate definite integrals ensures precise computations and a better grasp of the underlying mathematics.