Question

In the following integrals express the sines and cosines in exponential form and then integrate to show that: $$\int_{-\pi}^{\pi} \sin 2 x \sin 3 x d x=0$$

Short answer

The integral \[\int_{-\pi}^{\pi} \text{sin}(2x)\text{sin}(3x) dx = 0\].

Step-by-step solution

Convert Sine Functions to Exponential Form

Use Euler's formula to express \(\text{sin}(x) \) in exponential form: \[\text{sin}(x) = \frac{e^{ix} - e^{-ix}}{2i}\]. Therefore, \(\text{sin}(2x)\) and \(\text{sin}(3x)\) can be written as \[\text{sin}(2x) = \frac{e^{i2x} - e^{-i2x}}{2i}\] and \[\text{sin}(3x) = \frac{e^{i3x} - e^{-i3x}}{2i}\].

Multiply the Exponential Forms

To find the product \[\text{sin}(2x)\text{sin}(3x)\], multiply the exponential forms: \[\text{sin}(2x)\text{sin}(3x) = \left(\frac{e^{i2x} - e^{-i2x}}{2i}\right)\left(\frac{e^{i3x} - e^{-i3x}}{2i}\right)\].

Simplify the Product

Expand and simplify the product: \[\frac{(e^{i2x} - e^{-i2x})(e^{i3x} - e^{-i3x})}{(2i)(2i)} = \frac{1}{-4}\left(e^{i5x} - e^{ix} - e^{-ix} + e^{-i5x}\right)\].

Rewrite the Integrand

Rewrite the integrand to prepare for integration: \[\int_{-\pi}^{\pi} \text{sin}(2x)\text{sin}(3x) dx = \int_{-\pi}^{\pi} \frac{1}{-4}\left(e^{i5x} - e^{ix} - e^{-ix} + e^{-i5x}\right) dx\].

Integrate Each Term Separately

Integrate each term in the expression separately: \[\int_{-\pi}^{\pi} \frac{1}{-4}(e^{i5x} - e^{ix} - e^{-ix} + e^{-i5x}) dx\]. Each individual integral can be computed as follows: \[\frac{1}{-4}\left(\int_{-\pi}^{\pi} e^{i5x} dx - \int_{-\pi}^{\pi} e^{ix} dx - \int_{-\pi}^{\pi} e^{-ix} dx + \int_{-\pi}^{\pi} e^{-i5x} dx\right)\].

Evaluate Each Integral

Evaluate each integral with regard to the limits of integration. For any integer \ n eq 0 \ , \[\int_{-\pi}^{\pi} e^{inx} dx = 0\] because the integrals of purely imaginary exponentials over a full period cancel out.

Summarize the Results

Each term evaluates to 0: \[\frac{1}{-4}(0 - 0 - 0 + 0) = 0\]. Therefore, \[\int_{-\pi}^{\pi} \text{sin}(2x)\text{sin}(3x) dx = 0\].

Key concepts

Euler's formula

Euler's formula is an essential concept in trigonometry and complex analysis. It states that for any real number \( x \), the following holds true: \[ e^{ix} = \text{cos}(x) + i\text{sin}(x) \].
This formula bridges the gap between exponential functions and trigonometric functions.
By using Euler's formula, we can convert complex exponentials into sine and cosine functions, which makes calculations involving trigonometric integrals simpler.
The formula also showcases a deep connection between complex exponentials and circular motion, a principle used extensively in many fields of engineering and physics.

Exponential form of sine function

The exponential form of the sine function can be derived directly from Euler's formula.
Using Euler's formula, we get: \[ e^{ix} = \text{cos}(x) + i\text{sin}(x) \].
If we isolate \(\text{sin}(x)\), the formula becomes: \[ 2i \text{sin}(x) = e^{ix} - e^{-ix} \],
and thus we have: \[ \text{sin}(x) = \frac{e^{ix} - e^{-ix}}{2i} \].
This representation of the sine function in its exponential form is crucial when dealing with integrals, as it allows us to leverage properties of exponential functions which often simplify the integration process.

Integration of complex exponentials

Integrating complex exponentials often appears challenging, but becomes straightforward with some key properties.
For instance, consider the integral of an exponential function over a symmetric interval, such as \([-\pi, \pi]\): \[ \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{inx} dx \].
When \(n\) is an integer that is not zero, the integral evaluates to zero: \[ \int_{-\pi}^{\pi} e^{inx} dx = 0 \].
This happens because of the periodic nature of the exponential function; over one full period it sums to zero.
This property tremendously simplifies the integration of trigonometric functions expressed in their exponential form as it often leads terms to cancel out.

Orthogonality of sine functions

Orthogonality in mathematics refers to functions that integrate to zero when multiplied together over a specific interval.
In this context, sine functions of different frequencies exhibit orthogonality over a complete period \([-\pi, \pi]\).
For instance, consider the following result: \[ \int_{-\pi}^{\pi} \text{sin}(m x) \text{sin}(n x) dx = 0, \]
provided \(m \eq n\).
This orthogonality property greatly simplifies calculations involving such sine integrals, making it valuable in Fourier analysis and signal processing.
By expressing sine functions in their exponential form, the orthogonality becomes evident as the integral of the cross products of the exponentials cancel out, leaving us with zero.