Question

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

Short answer

\(\int_{-\pi}^{\pi} \sin 2 x \cos 3 x dx = 0\).

Step-by-step solution

- Express sin and cos in exponential form

Use Euler's formula to convert \(\sin 2x\) and \(\cos 3x\) into their exponential forms: \[\sin 2x = \frac{e^{i2x} - e^{-i2x}}{2i}\] and \[\cos 3x = \frac{e^{i3x} + e^{-i3x}}{2}.\]

- Multiply the exponential forms

Multiply the exponential forms of \(\sin 2x\) and \(\cos 3x\): \[\sin 2x \cos 3x = \left(\frac{e^{i2x} - e^{-i2x}}{2i}\right)\left(\frac{e^{i3x} + e^{-i3x}}{2}\right).\]

- Simplify the product

Expand and simplify the product of the exponentials: \[\sin 2x \cos 3x = \frac{1}{4i} \left(e^{i5x} + e^{-i5x} - e^{i x} - e^{-i x}\right).\]

- Set up the integral

Substitute the expression back into the integral: \[\int_{-\pi}^{\pi} \sin 2x \cos 3x \, dx = \int_{-\pi}^{\pi} \frac{1}{4i} \left(e^{i5x} + e^{-i5x} - e^{ix} - e^{-ix}\right) \, dx.\]

- Integrate term by term

Integrate each term separately: \[\int_{-\pi}^{\pi} e^{i5x} \, dx, \int_{-\pi}^{\pi} e^{-i5x} \, dx, \int_{-\pi}^{\pi} e^{ix} \, dx, \text{ and } \int_{-\pi}^{\pi} e^{-ix} \, dx.\] Each of these integrals evaluates to zero over the symmetric interval \([-\pi, \pi]\).

- Combine the results

Since each integral evaluates to zero, combining these results gives: \[\int_{-\pi}^{\pi} \sin 2x \cos 3x \, dx = 0.\]

Key concepts

Euler's formula

Euler's formula is a powerful tool in mathematics that relates complex exponentials to trigonometric functions. It states that \(e^{ix} = \text{cos}(x) + i\text{sin}(x)\). Another useful form is its conjugate \(e^{-ix} = \text{cos}(x) - i\text{sin}(x)\). With these formulas, we can represent sines and cosines in terms of exponential functions.
For example, sinus and cosine functions can be expressed as: \[ \text{sin}(x) = \frac{e^{ix} - e^{-ix}}{2i} \] and \[ \text{cos}(x) = \frac{e^{ix} + e^{-ix}}{2} \].
This conversion simplifies the manipulation and integration of trigonometric functions by transforming them into exponentials.

Trigonometric integrals

Trigonometric integrals often involve products of sine and cosine functions. These integrals can be difficult to handle directly. However, by converting the sine and cosine functions into their exponential forms using Euler's formula, we can simplify these integrals.
In the given exercise, we needed to integrate \(\text{sin}(2x)\text{cos}(3x)\) from \(-\pi\) to \(\pi\). This product can be expressed as: \[ \text{sin}(2x) \text{cos}(3x) = \left(\frac{e^{i2x} - e^{-i2x}}{2i}\right)\left(\frac{e^{i3x} + e^{-i3x}}{2}\right) \].
By transforming the product into an exponential form, it becomes easier to integrate term by term.

Complex exponentials

Complex exponentials are expressions involving \(e^{ix}\), where \(i\) is the imaginary unit defined as \(\sqrt{-1}\). These expressions are very useful in solving integrals and differential equations in which trigonometric functions are involved.
In our exercise, we dealt with exponentials like \(e^{i2x}\) and \(e^{i3x}\). When we performed multiplication and expanded the terms, we got: \[ \text{sin}(2x)\text{cos}(3x) = \frac{1}{4i} \left( e^{i5x} + e^{-i5x} - e^{ix} - e^{-ix} \right) \].
Each term represents a simple exponential function that's easier to integrate. This is a powerful technique to break down complex integrals into simpler forms.

Integration techniques

Integration is the process of finding the area under a curve. By expressing trigonometric functions as complex exponentials, as we have done using Euler's formula, the resulting integrals become simpler.
After converting our original integral \(\int_{-\pi}^{\pi} \text{sin}(2x)\text{cos}(3x)\, dx\), the integral transformed into: \[ \int_{-\pi}^{\pi} \text{sin}(2x) \text{cos}(3x) \, dx = \int_{-\pi}^{\pi} \frac{1}{4i} \left( e^{i5x} + e^{-i5x} - e^{ix} - e^{-ix} \right) dx \].
When integrating, each term like \(\int_{-\pi}^{\pi} e^{i5x} dx\) is found to be zero over the interval \([-\pi, \pi]\). This is due to the periodic nature of the exponential functions, thus simplifying our integral significantly to yield the final result of zero.