Question

∫x⁷(e^x)(sinx)dx

Answer 1

What a mean question, but one that will surely improve your skills at 'Integration by parts'

I've done integration by parts by first choosing:
u=x^7
du=7x^6 dx
dv = e^xsinx
v = ?

In order to take the integal of the v term you must perform Integration by Parts again.

\[\int\limits_{}^{}e^x Sinx dx\]
u = sinx
du = cosx
dv = e^x
v = e^x

Once you obtain this integral you can sub it into your original v term and perform one step of the integration by parts. Goodluck, this step only reduces your arithmetic term by one, meaning you have to do IBP another 6 times.

Answer 2

Hmm here's another idea also :D If you use Euler's Formula, you can write the sine function as the "imaginary part" of the complex exponential e^ix.

This leaves us with an integral involving only 2 terms, so it's easier to do by parts. You'll just need to pick out all the imaginary terms at the end because those are the ones we want.

Hmmmm either way, this problem is going to be a ton of work :c lol

\[\large \int\limits x^7 e^x \sin x dx=\Im\left(\int\limits x^7 e^x e^{ix}dx\right)=\Im\left(\int\limits x^7 e^{(1+i)x}dx\right)\]

Answer 3

\[\huge{sinx = \frac{e^{ix}-e^{-ix}}{2i}}\]
\[\huge{e^xsinx=e^x(\frac{e^{ix}-e^{-ix}}{2i})}\]
\[\huge{=\frac{1}{2i}x^7(e^{(1+i)x)}-e^{(1-i)x})}\]

Answer 4

Yah I wasn't using that identity :D thats useful also though ^^

I was just using this:
\[\huge e^{ix}=\cos x + i \sin x\]\[\huge Im(e^{ix})=\sin x\]
I'm not sure what the weird symbol is that I typed before :D I was just trying to type Im lol