integrate -x (sinx )(e^2x)

Mar 2014
909
2
malaysia
is it possible to integrate -x (sinx )(e^2x) ??
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I think I would try writing sin(x) as \(\displaystyle \frac{e^{ix}- e^{-ix}}{2i}\) so that \(\displaystyle -x sin(x) e^{2x}= -x \frac{e^{(2+i)x}- e^{(2- i)x}}{2i}\) and then use integration by parts on each of the two integrals.
 
Mar 2014
909
2
malaysia
You were shown that in reply #3. Why do you expect us to do it all?
i ahve never learnt to transform sin(x) as \(\displaystyle \frac{e^{ix}- e^{-ix}}{2i}\) , i prefer ordinary uv - ſv du
 

Plato

MHF Helper
Aug 2006
22,491
8,653
i ahve never learnt to transform sin(x) as \(\displaystyle \frac{e^{ix}- e^{-ix}}{2i}\) , i prefer ordinary uv - ſv du
If that is true, then why were you given this question? Whoever gave you this problem was irresponsible to assign a problem for which you do not have the necessary tools. Or did you just not study?
 
Mar 2014
909
2
malaysia
If that is true, then why were you given this question? Whoever gave you this problem was irresponsible to assign a problem for which you do not have the necessary tools. Or did you just not study?
So, the method aforementioned is the only way to solve this question??
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Yes it is possible to do using Integration by Parts, but it is a LOT of work!

First you will need to know what $\displaystyle \begin{align*} \int{ \mathrm{e}^{2\,x}\sin{(x)}\,\mathrm{d}x} \end{align*}$ is...

$\displaystyle \begin{align*} I &= \int{\mathrm{e}^{2\,x}\sin{(x)}\,\mathrm{d}x} \\ I &= \frac{1}{2}\,\mathrm{e}^{2\,x} \sin{(x)} - \int{ \frac{1}{2}\,\mathrm{e}^{2\,x}\cos{(x)}\,\mathrm{d}x} \\ I &= \frac{1}{2}\,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{2} \int{ \mathrm{e}^{2\,x}\cos{(x)}\,\mathrm{d}x } \\ I &= \frac{1}{2} \,\mathrm{e}^{2\,x}\sin{(x)} - \frac{1}{2}\, \left[ \frac{1}{2}\,\mathrm{e}^{2\,x} \cos{(x)} - \int{ -\frac{1}{2} \,\mathrm{e}^{2\,x}\sin{(x)}\,\mathrm{d}x} \right] \\ I &= \frac{1}{2} \,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{4}\,\mathrm{e}^{2\,x} \cos{(x)} - \frac{1}{4}\, I \\ \frac{5}{4}\,I &= \frac{1}{2}\,\mathrm{e}^{2\,x}\sin{(x)} - \frac{1}{4}\,\mathrm{e}^{2\,x}\cos{(x)} \\ I &= \frac{2}{5}\,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{5}\,\mathrm{e}^{2\,x} \cos{(x)} \end{align*}$

So that means $\displaystyle \begin{align*} \int{ \mathrm{e}^{2\,x} \sin{(x)}\,\mathrm{d}x} = \frac{2}{5}\,\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{5}\,\mathrm{e}^{2\,x} \cos{(x)} + C \end{align*}$. Now applying it to your problem $\displaystyle \begin{align*} \int{ -x\,\mathrm{e}^{2\,x} \sin{(x)}\,\mathrm{d}x} \end{align*}$ you need to use integration by parts with $\displaystyle \begin{align*} u = -x \implies \mathrm{d}u = -\mathrm{d}x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = \mathrm{e}^{2\,x} \sin{(x)} \,\mathrm{d}x \implies v = \frac{2}{5}\mathrm{e}^{2\,x} \sin{(x)} - \frac{1}{5}\,\mathrm{e}^{2\,x}\cos{(x)} \end{align*}$. Have a go...