Finding anitderivative using complex numbers and Euler

[Rectifier]

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.

Please help.

 

[phyzguy]

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}.

 

[Ray Vickson]

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.

Please help.

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.

 

[Ray Vickson]

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]

Thank you for your help