Question

integrate 1/4e^(3x) cos(2x)

Answer 1

\[\int\frac{1}{4}e^{3x}\cos(2x)~dx~?\]
Integrate by parts twice.

Answer 2

It's confusing me "/

Answer 3

For the first round of IBP, let
\[\begin{matrix}
u=e^{3x}&dv=\cos(2x)~dx\\
du=3e^{3x}~dx&v=\frac{1}{2}\sin(2x)
\end{matrix}\]
\[\begin{align*}\frac{1}{4}\int e^{3x}\cos(2x)~dx&=\frac{1}{4}\left[\frac{1}{2}e^{3x}\sin(2x)-\frac{3}{2}\int e^{3x}\sin(2x)~dx\right]\\
&=\frac{1}{8}e^{3x}\sin(2x)-\frac{3}{8}\int e^{3x}\sin(2x)~dx
\end{align*}\]
For the next round, let
\[\begin{matrix}
u=e^{3x}&dv=\sin(2x)~dx\\
du=3e^{3x}~dx&v=-\frac{1}{2}\cos(2x)
\end{matrix}\]
\[\begin{align*}\frac{1}{4}\color{red}{\int e^{3x}\cos(2x)~dx}&=\frac{1}{4}\left[\frac{1}{2}e^{3x}\sin(2x)-\frac{3}{2}\int e^{3x}\sin(2x)~dx\right]\\
&=\frac{1}{8}e^{3x}\sin(2x)-\frac{3}{8}\left[-\frac{1}{2}e^{3x}\cos(2x)+\frac{3}{2}\int e^{3x}\cos(2x)~dx\right]\\
&=\frac{1}{8}e^{3x}\sin(2x)+\frac{3}{16}e^{3x}\cos(2x)-\frac{9}{16}\color{red}{\int e^{3x}\cos(2x)~dx}
\end{align*}\]

Answer 4

Thank you so much :D That really helped me!!!

Answer 5

You're welcome!