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...