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Evaluate the two integrals by considering $C+iS$ $$C=\int_0^{\pi/2} e^{2x}\cos 3x dx \qquad S=\int_0^{\pi/2}e^{2x}\sin3xdx$$

So I have attempted this, but I have gone wrong with it somewhere so my answer is still a way from the correct answers which are :$$C=-\frac{1}{13}(2+3e^\pi) \qquad S=\frac{1}{13}(3-2e^\pi)$$ My workings for this are given as follows: $$C+iS = \int_0^{\pi/2}e^{2x}(\cos3x + i\sin3x)dx$$ $$\int_0^{\pi/2}e^{2x}(e^{3ix})dx$$ $$\int_0^{\pi/2}e^{(2+3i)x}dx = \left[ \frac{1}{2+3i}e^{(2+3i)x}\right]_0^{\pi/2}$$ $$ \frac{1}{2+3i}\left[ e^{(2+3i)(\pi/2)}-1\right] = \frac{1}{2+3i}\left[ e^\pi e^{3\pi i/2}-1\right]$$ $$\frac{e^\pi}{2+3i}\left[ \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}-1\right] = e^\pi\left( \frac {2}{13} - \frac{3i}{13} \right)\left( \cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) - e^\pi\left( \frac {2}{13} - \frac{3i}{13} \right)$$

This is kinda how far I got as I know I need to take the real and imaginary parts for each integral, but doing that gives me $$C=\frac{-5e^\pi}{13}$$ Which is wrong. Any help would be greatly appreciated.

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    $\begingroup$ When you factored $e^\pi$ out, you should change the $1$. $\endgroup$ – Clayton Apr 13 at 20:11
  • $\begingroup$ Ah yes, thank you. I probably should have spotted that $\endgroup$ – H.Linkhorn Apr 13 at 20:15
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Although a comment already cleared this up, let's make the calculation explicit using $\exp\frac{3\pi i}{2}=-i$: $$C+iS=-\frac{1+e^\pi i}{2+3i}=-\frac{(1+e^\pi i)(2-3i)}{13}=\frac{-2-3e^\pi + i(3-2e^\pi)}{13}.$$

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