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 \(\text{{\int}}_{{-\pi}}^{{\pi}} \sin 2x \sin 3x \text{{d}} x = 0\).

Step-by-step solution

- Express \sin 2x in exponential form

Using Euler's formula, \(\text{sin}(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i} \), we can write \(\text{sin}(2x)\) as: \[\text{sin}(2x) = \frac{e^{i2x} - e^{-i2x}}{2i} \]

- Express \sin 3x in exponential form

Similarly, we use Euler's formula to express \(\text{sin}(3x)\): \[\text{sin}(3x) = \frac{e^{i3x} - e^{-i3x}}{2i} \]

- Substitute sin terms into the integral

Substitute the exponential forms from steps 1 and 2 into the integral: \[ \int_{-\pi}^{\pi} \sin 2x \sin 3x \mathrm{d}x = \int_{-\pi}^{\pi} \left( \frac{e^{i2x} - e^{-i2x}}{2i} \right) \left( \frac{e^{i3x} - e^{-i3x}}{2i} \right) \mathrm{d}x \]

- Simplify the integrand

Simplify the product inside the integral: \[ \left( \frac{e^{i2x} - e^{-i2x}}{2i} \right) \left( \frac{e^{i3x} - e^{-i3x}}{2i} \right) = \frac{1}{4} \left( e^{i5x} - e^{-i5x} - e^{ix} + e^{-ix} \right) \]

- Split the integral

Now, we split the integral into four parts: \[\frac{1}{4} \int_{-\pi}^{\pi} \left( e^{i5x} - e^{-i5x} - e^{ix} + e^{-ix} \right) \mathrm{d}x \]

- Integrate each exponential term

Since all the terms are exponentials of the form \(e^{ikx}\), their integrals from \(-\pi\) to \(\pi\) can be evaluated: \[\frac{1}{4} \left( \int_{-\pi}^{\pi} e^{i5x} \mathrm{d}x - \int_{-\pi}^{\pi} e^{-i5x} \mathrm{d}x - \int_{-\pi}^{\pi} e^{ix} \mathrm{d}x + \int_{-\pi}^{\pi} e^{-ix} \mathrm{d}x \right) \] Each of these integrals evaluates to zero because \(e^{ikx}\) oscillates and its integral over a symmetric interval around zero is zero.

- Conclude the result

Since all four terms in the integrand integrate to zero, the integral evaluates to zero: \[ \int_{-\pi}^{\pi} \sin 2x \sin 3x \mathrm{d}x = 0 \]

Key concepts

Euler's Formula

Euler's formula is a crucial tool when working with trigonometric functions in complex analysis. It states that for a given angle \theta, the complex exponential function can be represented as:
\[\begin{equation} e^{i \theta} = \text{cos}(\theta) + i \text{sin}(\theta) \end{equation}\].

This formula connects the exponential function to the trigonometric functions sine and cosine, making it easier to manipulate and solve integrals involving trigonometric functions.

To express sine or cosine in exponential form, you can solve the formulas for \text{cos}(\theta) and \text{sin}(\theta) in terms of exponentials:

• \text{sin}(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}
• \text{cos}(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}

If we apply Euler's formula to the example integral, we start by expressing \text{sin(2x)} and \text{sin(3x)} in terms of exponential functions, which will simplify the integration process.

Trigonometric Integrals

Trigonometric integrals are those involving sine, cosine, and other trigonometric functions in the integrand. They often benefit from transformation into exponential form, utilizing Euler's formula.

Consider the integral given in the exercise:
<\[\begin{equation} \int_{-\pi}^{\pi} \sin(2x) \sin(3x) \mathrm{d}x \end{equation}\].

We'll convert \text{sin(2x)} and \text{sin(3x)} into their exponential forms:

• \text{sin(2x)} = \frac{e^{i2x} - e^{-i2x}}{2i}
• \text{sin(3x)} = \frac{e^{i3x} - e^{-i3x}}{2i}

Substituting these into the integral simplifies our calculations, allowing us to multiply the exponential expressions and integrate term-by-term.

To solve such integrals, understanding the properties of exponential functions is essential. Specifically, the integrals of exponential functions of the form \text{e^{ikx}} over a symmetric interval around zero will always evaluate to zero, due to the oscillatory nature of exponentials. This makes solving the integral straightforward.

Exponential Form

Using exponential form to represent trigonometric functions significantly eases the process of integrating products of trigonometric functions. Instead of dealing directly with sine and cosine, their exponential equivalents often lead to simpler manipulation and resolution.

For the integral in question, once expressed in exponential form, the process becomes a matter of integrating exponential terms. Let's break down the steps involved:

• First, substitute the exponential forms of the sines: \[\begin{equation} \int_{-\pi}^{\pi} \left( \frac{e^{i2x} - e^{-i2x}}{2i} \right) \left( \frac{e^{i3x} - e^{-i3x}}{2i} \right) \mathrm{d}x \end{equation}\].
• Next, simplify the product inside the integral: \[\begin{equation} \left( \frac{e^{i2x} - e^{-i2x}}{2i} \right) \left( \frac{e^{i3x} - e^{-i3x}}{2i} \right) = \frac{1}{4} \left( e^{i5x} - e^{-i5x} - e^{ix} + e^{-ix} \right) \end{equation}\].
• Then, split and integrate each term separately: \[\begin{equation} \frac{1}{4} \left( \int_{-\pi}^{\pi} e^{i5x} \mathrm{d}x - \int_{-\pi}^{\pi} e^{-i5x} \mathrm{d}x - \int_{-\pi}^{\pi} e^{ix} \mathrm{d}x + \int_{-\pi}^{\pi} e^{-ix} \mathrm{d}x \right) \end{equation}\].
• Finally, observe that each integral of the form \text{e^{ikx}} over \text{-\pi to \pi} evaluates to zero.

Conversion to exponential form and subsequent simplification caters well to these results. When utilizing exponential forms, always consider the symmetry and periodicity features of the trigonometric functions at play.