Question

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

Short answer

The integral evaluates to 0.

Step-by-step solution

Understanding Symmetry

The given integral is over a symmetric interval, from \(-\pi\) to \(\pi\), and we need to check if the function \(f(x) = \sin x + \cos x\) is symmetric. A function \(f(x)\) is considered even if \(f(-x) = f(x)\) and odd if \(f(-x) = -f(x)\).

Determine Symmetry Type

Let's evaluate \(f(-x) = \sin(-x) + \cos(-x)\). Using the identities \(\sin(-x) = -\sin(x)\) and \(\cos(-x) = \cos(x)\), we find \(f(-x) = -\sin(x) + \cos(x)\). Since this expression does not satisfy the conditions for being entirely even or odd, \(f(x)\) is neither even nor odd.

Splitting the Integral

Even though the entire function is not symmetric, we might benefit from splitting the integral into two parts: \[ \int_{-\pi}^{\pi} \sin x \, dx + \int_{-\pi}^{\pi} \cos x \, dx \] since \(\sin x\) is an odd and \(\cos x\) is an even function.

Evaluate the Integral of Sine

Since \(\sin x\) is an odd function, \(\int_{-a}^{a} \sin x \, dx = 0\) over any symmetric interval centered at zero. Therefore, \(\int_{-\pi}^{\pi} \sin x \, dx = 0\).

Evaluate the Integral of Cosine

The function \(\cos x\) is even, so evaluate \(\int_{-\pi}^{\pi} \cos x \, dx\): Use the property of even functions to get \(2 \int_{0}^{\pi} \cos x \, dx = 2 [\sin x]_{0}^{\pi}\). Calculating this, we find \(\sin \pi - \sin 0 = 0 - 0 = 0\), so \(\int_{-\pi}^{\pi} \cos x \, dx = 0\).

Combine Results

Add the results of the two integrals: \(\int_{-\pi}^{\pi} \sin x \, dx + \int_{-\pi}^{\pi} \cos x \, dx = 0 + 0 = 0\). Therefore, the integral \(\int_{-\pi}^{\pi}(\sin x + \cos x) \, dx\) evaluates to 0.

Key concepts

Symmetry of Functions

Symmetry plays a crucial role in simplifying definite integrals. When a function is symmetric, it gives us a great way to potentially reduce the complexity of an integral. There are two main forms of symmetry we look at in calculus: even symmetry and odd symmetry.
Consider a function evaluated over an interval that is symmetric about the origin, such as [-a, a]. If a function is even, this means that the function is mirrored over the y-axis. Conversely, if it is odd, it reflects over the origin.
This symmetry is not merely a matter of visual aesthetics; it influences the area under the curve and drastically affects how we evaluate integrals.
In our exercise, symmetry helps us identify that although the overall function is neither odd nor even, individual components of it might be, which allows us to split and evaluate simpler parts.

Even and Odd Functions

Understanding even and odd functions is key to determining symmetries involved in integrals. An even function satisfies the property that for every value of \( x \), \( f(-x) = f(x) \). A classic example is \( \cos x \). This property means that the graph of the function is mirrored along the y-axis.
Odd functions, on the other hand, follow the property \( f(-x) = -f(x) \). The sine function, \( \sin x \), is commonly cited as an odd function. This results in the graph being symmetric with respect to the origin.
In the given integral, \( \sin x + \cos x \), we can notice that upon evaluating \( f(-x) \), neither the even nor odd condition holds true. However, by isolating \( \sin x \) (odd) and \( \cos x \) (even), we can leverage the defining integrable properties of even and odd functions, simplifying the evaluation process.

Definite Integral Evaluation

When evaluating a definite integral, especially over symmetric limits like \([-\pi, \pi]\), symmetry can help us significantly. This interval can be split into simpler parts by recognizing the nature of the function included.

• For odd functions like \( \sin x \), integrals over symmetric intervals equate to zero. \( \int_{-a}^{a} \sin x \, dx = 0 \).
• For even functions such as \( \cos x \), the integral from \( -a \) to \( a \) can be calculated by doubling the integral from 0 to \( a \), because of its action under reflection.

By employing these properties, evaluations become straightforward. In our example, both \( \int_{-\pi}^{\pi} \sin x \, dx \) and \( \int_{-\pi}^{\pi} \cos x \, dx \) result in zero, thus reducing our expression smoothly.

Trigonometric Integrals

Trigonometric integrals often pose complex challenges but also reveal elegant solutions when symmetry and function properties are exploited. Sinusoidal functions like sine and cosine have inherent symmetrical qualities which simplify calculations.
A deeper look into their integral evaluation: \( \sin x \), being odd, cancels itself out over symmetric limits, while \( \cos x \), being even, does not inherently yield a non-zero result when evaluated. Knowledge of these intricacies is not only useful but necessary in solving integrals involving trigonometric functions.
Understanding the nature of \( \sin x \) and \( \cos x \) pays off significantly in calculus, allowing for effective problem-solving and comprehension. For practical integration over a symmetric interval, one should always verify the function's properties to see if these simplifying rules can be applied.