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

#### xl5899

is it possible to integrate -x (sinx )(e^2x) ??

MHF Helper

#### HallsofIvy

MHF Helper
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.

#### xl5899

According to Wolfram Alpha it is ...
can someone show the full working? i need it . i tried to do this many times , but i did a lot of mistake halfway

#### Plato

MHF Helper
can someone show the full working? i need it .
You were shown that in reply #3. Why do you expect us to do it all?

#### xl5899

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
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?

#### xl5899

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??

MHF Helper

#### Prove It

MHF Helper
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...