Question

int [xe^xsinx]dx =?

Answer 1

\[\int\limits xe^xsinxdx =?\]

Answer 2

integration by parts...have you tried it

Answer 3

Do you know Integration by parts??? @cahit

Answer 4

yes i used u=X and the other dv but i couldnt solve...
or it is so long...

Answer 5

if i know wich part is U then i solve it easily..

Answer 6

u= x ; dv = e^(x)sin(x)
use integration by parts for ∫e^(x)sin(x) first

Answer 7

or use
http://en.wikipedia.org/wiki/Integration_using_Euler%27s_formula

Answer 8

Here is it worked out in case you're still stuck :)

\[\int x\mbox{e}^{x}\sin x\mbox{ d}x\]

Let \(u=x\) and \(\mbox{d}v=\mbox{e}^{x}\sin x\mbox{ d}x\), then \(\mbox{d}u=\mbox{d}x\) and \(v=\int e^{x}\sin x\mbox{ d}x\)


\[v=\int e^{x}\sin x\mbox{ d}x\]
let \(u=\mbox{e}^{x}\) and \(\mbox{d}v=\sin x\mbox{ d}x\), \(\mbox{d}u=\mbox{e}^{x}\mbox{d}x\), \(v=-\cos x\)


\[\int e^{x}\sin x\mbox{ d}x=-\mbox{e}^{x}\cos x+\int\mbox{e}^{x}\cos x\mbox{ d}x\]

for the integral \(\int\mbox{e}^{x}\cos x\mbox{ d}x\), let \(u=\mbox{e}^{x}\) and \(\mbox{d}v=\cos x\mbox{ d}x\), so \(\mbox{d}u=\mbox{e}^{x}\mbox{d}x\), \(v=\sin x\)


\[\int\mbox{e}^{x}\sin x\mbox{ d}x=-\mbox{e}^{x}\cos x+\mbox{e}^{x}\sin x-\int\mbox{e}^{x}\sin x\mbox{ d}x\]

\[2\int\mbox{e}^{x}\sin x\mbox{ d}x=\mbox{e}^{x}\left(\sin x-\cos x\right)\]
\[\int\mbox{e}^{x}\sin x\mbox{ d}x=\frac{1}{2}\mbox{e}^{x}\left(\sin x-\cos x\right)\]


now we have \(v=\frac{1}{2}\mbox{e}^{x}\left(\sin x-\cos x\right)\), so

\[\int x\mbox{e}^{x}\sin x\mbox{ d}x = \frac{x}{2}\mbox{e}^{x}\left(\sin x-\cos x\right)-\frac{1}{2}\int\mbox{e}^{x}\left(\sin x-\cos x\right)\mbox{d}x =\]
\[\frac{x}{2}\mbox{e}^{x}\left(\sin x-\cos x\right)-\frac{1}{2}\int\mbox{e}^{x}\sin x\mbox{ d}x+\frac{1}{2}\int\mbox{e}^{x}\cos x\mbox{ dx}\]


we have already found \(\int\mbox{e}^{x}\sin x\mbox{ d}x\) (it was v), so all that's left to do is the integral \(\int\mbox{e}^{x}\cos x\mbox{ d}x\), which is a similar process.
Let \(u=\mbox{e}^{x}\), \(\mbox{d}u=\mbox{e}^{x}\mbox{d}x\), \(\mbox{d}v=\cos x\mbox{ d}x\), \(v=\sin x\)


\[\int\mbox{e}^{x}\cos x\mbox{ d}x=\mbox{e}^{x}\sin x-\int\mbox{e}^{x}\sin x\mbox{ d}x\]


Since \(\int\mbox{e}^{x}\sin x\mbox{ d}x=\frac{1}{2}\mbox{e}^{x}\left(\sin x-\cos x\right)\)
\[\int\mbox{e}^{x}\cos x\mbox{ d}x=\frac{1}{2}\mbox{e}^{x}\sin x+\frac{1}{2}\mbox{e}^{x}\cos x=\frac{1}{2}\mbox{e}^{x}\left(\sin x+\cos x\right)\]


Substituting into our original integral we get

\[\int x\mbox{e}^{x}\sin x\mbox{ d}x = \frac{x}{2}\mbox{e}^{x}\left(\sin x-\cos x\right)-\frac{1}{4}\mbox{e}^{x}\left(\sin x-\cos x\right)+\frac{1}{4}\mbox{e}^{x}\left(\sin x+\cos x\right) =\]
\[\frac{1}{2}x\mbox{e}^{x}\sin x-\frac{1}{2}x\mbox{e}^{x}\cos x-\frac{1}{4}\mbox{e}^{x}\sin x+\frac{1}{4}\mbox{e}^{x}\cos x+\frac{1}{4}\mbox{e}^{x}\sin x+\frac{1}{4}\mbox{e}^{x}\cos x = \]
\[\frac{1}{2}\mbox{xe}^{x}\sin x-\frac{1}{2}x\mbox{e}^{x}\cos x+\frac{1}{2}\mbox{e}^{x}\cos x=\]
\[\frac{1}{2}\mbox{e}^{x}\left(x\sin x-x\cos x+\cos x\right)\]