Question

In the following integrals express the sines and cosines in exponential form and then integrate to show that: $$\int_{-\pi}^{\pi} \sin 3 x \cos 4 x d x=0$$

Short answer

\( \int_{-\pi}^{\pi} \sin 3 x \cos 4 x dx = 0 \)

Step-by-step solution

Express Sine and Cosine in Exponential Form

Use the Euler's formulas to express \(\sin 3x\) and \(\cos 4x\) in exponential form: \[ \sin 3x = \frac{e^{i 3x} - e^{-i 3x}}{2i} \] \[ \cos 4x = \frac{e^{i 4x} + e^{-i 4x}}{2} \]

Multiply the Exponential Forms

Multiply the expressions for \( \sin 3x \) and \( \cos 4x \) together: \[ \sin 3x \cos 4x = \left(\frac{e^{i 3x} - e^{-i 3x}}{2i}\right) \left(\frac{e^{i 4x} + e^{-i 4x}}{2}\right) \]

Simplify the Product

Expand and simplify the product: \[ \sin 3x \cos 4x = \frac{1}{4i} (e^{i(3x + 4x)} + e^{i(3x - 4x)} - e^{-i(3x + 4x)} - e^{-i(3x - 4x)}) = \frac{1}{4i} (e^{i 7x} + e^{-i x} - e^{i x} - e^{-i 7x}) \]

Set Up the Integral

Set up the integral with the simplified form: \[ \int_{-\pi}^{\pi} \sin 3x \cos 4x dx = \int_{-\pi}^{\pi} \frac{1}{4i} (e^{i 7x} + e^{-i x} - e^{i x} - e^{-i 7x}) dx \]

Integrate Each Term Separately

Integral of each term separately: \[ \int_{-\pi}^{\pi} e^{i 7x} dx, \quad \int_{-\pi}^{\pi} e^{-i x} dx, \quad \int_{-\pi}^{\pi} e^{i x} dx, \quad \int_{-\pi}^{\pi} e^{-i 7x} dx \] Each of these integrals evaluates to zero because the integrands are complex exponentials with nonzero frequencies over a symmetric interval around zero.

Sum the Results

Since each integral evaluates to zero, sum the results: \[ \frac{1}{4i} (0 + 0 - 0 - 0) = 0 \]

Key concepts

Euler's formula

Euler's formula is a fundamental equation in complex analysis that relates complex exponentials to trigonometric functions. It's given by \[e^{ix} = \text{cos} \, x + i \, \text{sin} \, x\].
This formula is incredibly useful for converting sines and cosines into exponential forms, simplifying integration and other operations. In the given problem, we use Euler's formula to express \( \text{sin} \, 3x \) and \( \text{cos} \, 4x \) in terms of complex exponentials. This allows us to manage the problem with the powerful tools from complex analysis.
By expressing trigonometric functions in exponential form, the integrand becomes a sum of complex exponentials. This approach leverages the orthogonality properties of exponentials to simplify the integral.

Exponential form

In mathematics, expressing trigonometric functions in their exponential form, as derived from Euler's formula, can simplify many problems. The exponential form of sine and cosine are:
\[ \text{sin} \, 3x = \frac{e^{i 3x} - e^{-i 3x}}{2i}, \quad \text{cos} \, 4x = \frac{e^{i 4x} + e^{-i 4x}}{2}\].
Using these, the product \( \text{sin} \, 3x \, \text{cos} \, 4x \) is transformed into a more manageable form involving complex exponentials. These expressions can be easily manipulated, allowing for straightforward integration.
When converting to exponential form, the trigonometric identities and properties like periodicity and orthogonality become more apparent, streamlining the solution process.

Complex exponentials

Complex exponentials, expressed as \(e^{ix}\), are essential in simplifying many mathematical operations, including integration. They reveal the underlying symmetrical properties of the functions.
The orthogonal nature of complex exponentials over symmetric intervals, such as \([-\pi, \pi]\), means that integrals of such functions with different frequencies often result in zero.
This orthogonality property was used in the given problem where we integrated each term of the sum separately, showing that each integral of the form \( \frac{1}{4i} \int_{-\pi}^{\pi} e^{i kx} \, dx \) (where \(k \, eq \, 0\)) evaluates to zero.
Breaking down the sin and cos into complex exponentials made it easier to see why each resulting integral evaluates to zero, without extensive calculations.

Definite integrals

A definite integral calculates the net area under a curve within a specific interval. It is noted as:\[ \int_{a}^{b} f(x) \, dx \].
In this problem, we evaluated the definite integral of a product of trigonometric functions over the interval \([-\pi, \pi]\). By transforming the trigonometric functions into their exponential form, we turned the problem into integrating sums of complex exponentials.
This approach simplified our calculations. Each integral of the form \( \int_{-\pi}^{\pi} e^{i kx} \, dx \) where \(k \, eq \, 0\) turned out to be zero because of the orthogonality property of exponentials. Thus, the entire integral was zero.

Orthogonality

Orthogonality in functions implies that over a specific interval, the inner product (integral) of the functions equals zero unless the functions are identical (and normalized). For complex exponentials, this orthogonality is expressed as:\[ \int_{-\pi}^{\pi} e^{i kx} \, dx = 0, \quad \text{for any } k \, eq \, 0 \].
This principle is what makes the integral of \( \text{sin} \, 3x \, \text{cos} \, 4x \) equal to zero in the given problem. When we decomposed \( \text{sin} \, 3x \) and \( \text{cos} \, 4x \) into complex exponentials, we ended up integrating terms of the form \( e^{i kx} \). These integrals over the symmetric interval \([-\pi, \pi]\) are zero unless the exponent is zero, simplifying our task significantly.