# Finding anitderivative using complex numbers and Euler

• Rectifier
In summary, the problem is to find a primitive function using the Euler formulas for sine and cosine. The attempt involves using the relevant equations and integrating, as well as converting to polar form for certain values. The next step is to make the denominators of the two expressions real and then use elementary algebra to express the result as a complex number.
Rectifier
Gold Member
I have to find a primitive function below using the Euler formulas for ##\sin x## and ## \cos x##

The problem

$$\int e^{2x} \sin 3x \ dx$$

Relevant equations

## \cos x = \frac{e^{ix}+e^{-ix}}{2} \\ \sin x = \frac{e^{ix}-e^{-ix}}{2i} \\ \\ \int e^{ix} \ dx = \frac{e^{ix}}{i} ##

The attempt
## \int e^{2x} \sin 3x \ dx = \int e^{2x} \left( \frac{ e^{i3x}-e^{-i3x} }{ 2i } \right) \ dx = \frac{ 1 }{ 2i } \int e^{ 2x }e^{i3x}-e^{2x}e^{-i3x} \ dx = \frac{1}{2i} \int e^{(2+3i)x}-e^{(2-3i)x} \ dx \\ \frac{1}{2i} \int e^{(2+3i)x}-e^{(2-3i)x} \ dx = \frac{1}{2i} \left( \frac{e^{(2+3i)x}}{2+3i}- \frac{e^{(2-3i)x}}{2-3i} \right) \ + C ##

Polar form for ##z=2+i##
##2+3i = \sqrt{13}e^{i \arctan \left( \frac{3}{2} \right) }##

Polar form for ##z=2-i##
##2-3i = \sqrt{13}e^{- i \arctan \left( \frac{3}{2} \right) }##I am looking for a way to advance from this step in my solution process and am therefore not interested in any alternative ways other than using Eulers forumlas above to solve this problem.

It looks good so far. After doing the integration, the next step should be to make the denominators of the two expressions real so that you can get to a common denominator and add them together. So multiply the first expression by $\frac{2-3i}{2-3i}$, and multiply the second expression by $\frac{2+3i}{2+3i}$.

Rectifier
Rectifier said:
I have to find a primitive function below using the Euler formulas for ##\sin x## and ## \cos x##

The problem

$$\int e^{2x} \sin 3x \ dx$$

Relevant equations

## \cos x = \frac{e^{ix}+e^{-ix}}{2} \\ \sin x = \frac{e^{ix}-e^{-ix}}{2i} \\ \\ \int e^{ix} \ dx = \frac{e^{ix}}{i} ##

The attempt
## \int e^{2x} \sin 3x \ dx = \int e^{2x} \left( \frac{ e^{i3x}-e^{-i3x} }{ 2i } \right) \ dx = \frac{ 1 }{ 2i } \int e^{ 2x }e^{i3x}-e^{2x}e^{-i3x} \ dx = \frac{1}{2i} \int e^{(2+3i)x}-e^{(2-3i)x} \ dx \\ \frac{1}{2i} \int e^{(2+3i)x}-e^{(2-3i)x} \ dx = \frac{1}{2i} \left( \frac{e^{(2+3i)x}}{2+3i}- \frac{e^{(2-3i)x}}{2-3i} \right) \ + C ##

Polar form for ##z=2+i##
##2+3i = \sqrt{13}e^{i \arctan \left( \frac{3}{2} \right) }##

Polar form for ##z=2-i##
##2-3i = \sqrt{13}e^{- i \arctan \left( \frac{3}{2} \right) }##I am looking for a way to advance from this step in my solution process and am therefore not interested in any alternative ways other than using Eulers forumlas above to solve this problem.

Do NOT express ##2 + 3i## in polar form; just use ##e^{(2 + 3i)x} = e^{2x} e^{3ix}## once again after you have done the integration. Using the polar form where it is not needed just makes everything worse.

Rectifier
When I do that, I get

## \frac{1}{2i} \left( \frac{ e^{2x} e^{3ix} (2-3i)}{(2+3i)(2-3i)}- \frac{e^{2x}e^{-3ix}(2+3i)}{(2+3i)(2-3i)} \right) \ + C = \frac{1}{2i} \left( \frac{ e^{2x} e^{3ix} (2-3i) - e^{2x}e^{-3ix}(2+3i)}{13}\right) \ + C \\ = \frac{e^{2x}}{2i} \left( \frac{ e^{3ix} (2-3i) - e^{-3ix}(2+3i)}{13}\right) \ + C ##

I get stuck here.

Rectifier said:
When I do that, I get

## \frac{1}{2i} \left( \frac{ e^{2x} e^{3ix} (2-3i)}{(2+3i)(2-3i)}- \frac{e^{2x}e^{-3ix}(2+3i)}{(2+3i)(2-3i)} \right) \ + C = \frac{1}{2i} \left( \frac{ e^{2x} e^{3ix} (2-3i) - e^{2x}e^{-3ix}(2+3i)}{13}\right) \ + C \\ = \frac{e^{2x}}{2i} \left( \frac{ e^{3ix} (2-3i) - e^{-3ix}(2+3i)}{13}\right) \ + C ##

I get stuck here.

Just use elementary algebra to express ##(a +ib) \times (c + id)## as ## U +iV##; there are standard rules for doing that. Of course, you need to remember that ##e^{\pm 3ix} = \cos(3x) \pm i \sin(3x).##

Rectifier

## 1. What is the purpose of using complex numbers and Euler when finding an antiderivative?

The use of complex numbers and Euler allows for a more efficient and concise method of finding antiderivatives. It also expands the range of functions that can be integrated, as some functions with real coefficients may have complex antiderivatives.

## 2. How do complex numbers and Euler relate to finding antiderivatives?

Complex numbers and Euler are used to represent functions that cannot be expressed solely in terms of real numbers. By using Euler's formula, e^(ix) = cos(x) + i*sin(x), complex numbers can be separated into their real and imaginary components, making it easier to integrate.

## 3. What are the key steps in finding an antiderivative using complex numbers and Euler?

The key steps involve converting the function into its complex representation, separating it into real and imaginary parts, integrating each part separately, and then combining them back into a complex antiderivative.

## 4. Are there any limitations to using complex numbers and Euler when finding antiderivatives?

Yes, there are certain functions that cannot be integrated using this method, such as functions with singularities or branch points. It is important to check for these limitations before attempting to find an antiderivative using complex numbers and Euler.

## 5. How is finding an antiderivative using complex numbers and Euler different from the traditional method?

The traditional method of finding antiderivatives involves using techniques such as substitution or integration by parts. Using complex numbers and Euler is a more specialized method that can be used for specific types of functions, particularly those with complex coefficients or exponential terms.

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