- #1

Rectifier

Gold Member

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- 4

The problem

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.