Question

Integrate the following: INT e^2x cos3xdx

Answer 1

I'll teach you a new technique. Its called complexifying the integral.
First you must know that:
\[e^{ix}=\cos(x)+i \sin(x)\]
Where if I said:
\[\Re[e^{ix}]=\cos(x) ; \Im[e^{ix}]=\sin(x)\]
With that being said observe that:
\[e^{2x+3ix}=e^{2x}e^{3\xi}=e^{2x}(\cos(3x)+i \sin(3x)); \Re=e^{2x}\cos(3x)\]
So that means the integral must be the real part of the complex number above^.
So the integral is the same as:
\[\Re \left[ \int\limits e^{(2+3i)x} dx \right]=\Re \left[ \frac{e^{(2+3i)x}}{2+3i} \right]\]
Multiply the top and bottom by the complex conjugate and expand out the e^(whatever)
This gives:
\[\Re \left[ \frac{(e^{2x}\cos(3x)+e^{2x}\sin(3x))(2-3i)}{(2+3i)(2-3i)} \right]\]

Answer 2

From here multiply out and find the real part. Multiplying gives:
\[\Re \left[ \frac{1}{13} [ e^{2x}(2\cos(3x)+\sin(3x))+e^{2x}(2\sin(3x)-3\cos(3x))] \right]=\frac{e^{2x}}{13}(2\cos(3x)+\sin(3x))+C\]

Answer 3

After sin(3x) on the last line there should be a +C.

Answer 4

I hope this isn't too "complex" -.-

Answer 5

Also, there should be a 3 in front of the sin(3x) in the solution. I just mistyped again.

Answer 6

What do you think outkast?