Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use symmetry to help you evaluate the given integral. $$ \int_{-\pi}^{\pi}(\sin x+\cos x)^{2} d x $$

Step-by-Step Solution

Answer

The value of the integral is \(2\pi\).

Step 1: Expand the integrand

We start by expanding the integrand \((\sin x + \cos x)^2\). This expands to:\[(\sin x + \cos x)^2 = \sin^2 x + 2\sin x\cos x + \cos^2 x\]

Step 2: Simplify using trigonometric identities

Apply the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). Replace \(\sin^2 x + \cos^2 x\) in the expression:\[1 + 2\sin x \cos x\]

Step 3: Recognize symmetry and simplify the integral

The integral can be split into two parts:\[\int_{-\pi}^{\pi} 1 \, dx + \int_{-\pi}^{\pi} 2\sin x \cos x \, dx\]Notice that the second integral \(\int_{-\pi}^{\pi} \sin x \cos x \, dx\) is an odd function because \(\sin x \cos x\) changes sign from \(-x\) to \(x\), making its integral over symmetric bounds zero.

Step 4: Compute each integral separately

Now, 1. Compute the integral of the constant function 1:\[\int_{-\pi}^{\pi} 1 \, dx = [x]_{-\pi}^{\pi} = \pi - (-\pi) = 2\pi\]2. Compute the integral of the odd function (which we already found to be zero):\[\int_{-\pi}^{\pi} 2\sin x \cos x \, dx = 2 \cdot 0 = 0\]

Step 5: Calculate the final result

Add the results from each integral:\[2\pi + 0 = 2\pi\]

Key Concepts

Symmetry in Integrals

In integral calculus, symmetry can significantly simplify the computation of definite integrals. Symmetry properties are particularly useful when working with integrals over symmetric limits, such as \(-a\) to \(a\). One common type of symmetry is even and odd functions:

• **Even Functions**: Symmetrical about the y-axis. For these, \(f(x) = f(-x)\), and the integral from \(-a\) to \(a\) simplifies to \(-a\) to \(a\).
• **Odd Functions**: Symmetrical about the origin. For these, \(f(-x) = -f(x)\), and the integral from \(-a\) to \(a\) is zero.

In the given exercise, we encountered \(2 \sin x \cos x\), which is an odd function because \(\sin x\cos x\) changes sign when \(x\) is replaced with \(-x\), i.e., \(\sin(-x)\cos(-x) = -\sin x\cos x\). Hence, over symmetric limits, this integral evaluates to zero, simplifying the calculation dramatically.

Trigonometric Identities

Trigonometric identities are tools that allow us to manipulate and simplify expressions involving trigonometric functions, converting them into more manageable forms. They can simplify integration by reducing complex expressions into simpler terms or constants.
Some key trigonometric identities include:

• **Pythagorean Identity**: \(\sin^2 x + \cos^2 x = 1\). This identity is pivotal in rewriting the expression \(\sin^2 x + \cos^2 x\) as 1, simplifying our problem.
• **Double Angle Identities**: \(2 \sin x \cos x = \sin 2x\). This may occasionally help in converting products of sines and cosines into simpler expressions.

In the exercise, the Pythagorean identity helped us condense the expanded form \(\sin^2 x + 2\sin x\cos x + \cos^2 x\) into \(1 + 2\sin x\cos x\). This simplification was crucial for recognizing the odd function inherent in the expression.

Definite Integrals

Definite integrals calculate the net area under a curve over a specified interval. In the context of this exercise, we are evaluating a definite integral from \(-\pi\) to \(\pi\). Definite integrals have specific properties and steps:

• **Evaluate Limits**: Specify the lower and upper bounds of the integral.
• **Simplify Integrand**: If possible, simplify the expression within the integral. Use identities or symmetry.
• **Compute Integral**: Calculate each part separately if breaking it into sums or differences helps.

After simplified using trigonometric identities and symmetry, we found ourselves with two tasks: evaluating the integral of a constant \(\int_{-\pi}^{\pi} 1 \, dx\) and recognizing the integral of the odd function \(\int_{-\pi}^{\pi} 2\sin x\cos x \, dx\) was zero. The first integral resulted in \(2\pi\) as it represents the length of the interval \(-\pi\) to \(\pi\), while the second vanished due to symmetry properties.