evaluate ∫ ((e^(3ix)-e^(-3ix))/(2i))((e^(4ix)-e^(-4ix))/(2i)) dx

To evaluate the integral 
\[ \int \left( \frac{e^{3ix} - e^{-3ix}}{2i} \right) \left( \frac{e^{4ix} - e^{-4ix}}{2i} \right) \, dx, \]
we will follow these steps:

1. **Multiply the terms inside the integral:**

\[
\left( \frac{e^{3ix} - e^{-3ix}}{2i} \right) \left( \frac{e^{4ix} - e^{-4ix}}{2i} \right) = \frac{1}{(2i)^2} \left( e^{3ix} - e^{-3ix} \right) \left( e^{4ix} - e^{-4ix} \right).
\]

2. **Simplify the product:**

\[
\frac{1}{(2i)^2} \left( e^{3ix} - e^{-3ix} \right) \left( e^{4ix} - e^{-4ix} \right) = \frac{1}{-4} \left( e^{3ix}e^{4ix} - e^{3ix}e^{-4ix} - e^{-3ix}e^{4ix} + e^{-3ix}e^{-4ix} \right).
\]

\[
= -\frac{1}{4} \left( e^{7ix} - e^{-ix} - e^{ix} + e^{-7ix} \right).
\]

3. **Rewrite the integral with the simplified expression:**

\[
\int -\frac{1}{4} \left( e^{7ix} - e^{-ix} - e^{ix} + e^{-7ix} \right) \, dx.
\]

4. **Separate the integral into four parts:**

\[
= -\frac{1}{4} \left( \int e^{7ix} \, dx - \int e^{-ix} \, dx - \int e^{ix} \, dx + \int e^{-7ix} \, dx \right).
\]

5. **Integrate each exponential function:**

\[
\int e^{aix} \, dx = \frac{e^{aix}}{ai},
\]

where \( a \) is a constant.

\[
\int e^{7ix} \, dx = \frac{e^{7ix}}{7i},
\]

\[
\int e^{-ix} \, dx = \frac{e^{-ix}}{-i} = -\frac{e^{-ix}}{i},
\]

\[
\int e^{ix} \, dx = \frac{e^{ix}}{i},
\]

\[
\int e^{-7ix} \, dx = \frac{e^{-7ix}}{-7i} = -\frac{e^{-7ix}}{7i}.
\]

6. **Substitute the results back into the integral:**

\[
= -\frac{1}{4} \left( \frac{e^{7ix}}{7i} - \left( -\frac{e^{-ix}}{i} \right) - \frac{e^{ix}}{i} + \left( -\frac{e^{-7ix}}{7i} \right) \right).
\]

7. **Simplify the expression:**

\[
= -\frac{1}{4} \left( \frac{e^{7ix}}{7i} + \frac{e^{-ix}}{i} - \frac{e^{ix}}{i} - \frac{e^{-7ix}}{7i} \right).
\]

Factor out \(\frac{1}{i}\):

\[
= -\frac{1}{4i} \left( \frac{e^{7ix}}{7} + e^{-ix} - e^{ix} - \frac{e^{-7ix}}{7} \right).
\]

Combine the constants:

\[
= -\frac{1}{4i} \left( \frac{e^{7ix}}{7} - \frac{e^{-7ix}}{7} + e^{-ix} - e^{ix} \right).
\]

8. **Express the result:**

\[
= -\frac{1}{4i} \left( \frac{e^{7ix} - e^{-7ix}}{7} + e^{-ix} - e^{ix} \right).
\]

Recognize that \( e^{aix} - e^{-aix} = 2i \sin(ax) \):

\[
= -\frac{1}{4i} \left( \frac{2i \sin(7x)}{7} + 2i \sin(-x) \right).
\]

\[
= -\frac{1}{4i} \left( \frac{2i \sin(7x)}{7} - 2i \sin(x) \right).
\]

\[
= -\frac{1}{4i} \cdot 2i \left( \frac{\sin(7x)}{7} - \sin(x) \right).
\]

The \(2i\) terms cancel out:

\[
= -\frac{1}{2} \left( \frac{\sin(7x)}{7} - \sin(x) \right).
\]

So, the integral is:

\[
\boxed{\int \left( \frac{e^{3ix} - e^{-3ix}}{2i} \right) \left( \frac{e^{4ix} - e^{-4ix}}{2i} \right) \, dx = -\frac{1}{14} \sin(7x) + \frac{1}{2} \sin(x) + C.}
\]

Would you like more details or have any questions about this process? Here are some related questions you might want to explore:

1. How can you use trigonometric identities to solve integrals involving products of sine and cosine?
2. What are the steps for integrating functions using partial fractions?
3. How can you derive Euler's formula from the Taylor series?
4. What are the methods for evaluating integrals using substitution?
5. How do you integrate functions involving hyperbolic sine and cosine?

**Tip:** When integrating products of trigonometric functions, converting them to exponential form using Euler's formula can simplify the process, especially for complex integrals.

Learn how to evaluate the integral of ((e^(3ix) - e^(-3ix))/(2i))((e^(4ix) - e^(-4ix))/(2i)) dx step-by-step. This problem involves complex numbers, exponential functions, and trigonometric identities. Suitable for advanced undergraduate students and those interested in complex analysis.

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