1. ## please derive this integral.. thanks a lot!

2. You should upload it on the internet then use the img tags for the URL of the image. Or you can just attach it here.

3. ## ...

Originally Posted by Chop Suey
You should upload it on the internet then use the img tags for the URL of the image. Or you can just attach it here.

sorry!! i edited it..
i really need your help!! thanks a lot!

4. to avoid confusion, i changed u to x
Originally Posted by vicky0514
integral of e^(au)sin(nu)du=(((e^au)(asinnu-nconnu))/(a^2+n^2))+C
do you mean $\int e^{ax} \sin nx~dx = \frac {e^{ax} (a \sin nx - n \cos nx)}{a^2 + n^2} + C$ ?

try integration by parts. let $u = \sin nx$ and $dv = e^{ax}~dx$. (i hope you remember what $u$ and $dv$ mean)

if you are really lazy, you can do a search for this question. it has been derived several times on this forum and many times on the internet, i'm sure, using both integration by parts and complex analysis

5. Hello,

Substitute : $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i} \implies \sin(nx)=\frac{1}{2i} \cdot (e^{inx}-e^{-inx})$

$\int e^{ax} \sin(nx) dx=\frac{1}{2i} \cdot \left(\int e^{ax}e^{inx} dx-\int e^{ax}e^{-inx} dx\right)$ $=\frac{1}{2i} \cdot \left(\int e^{x(a+in)} dx-\int e^{x(a-in)} dx \right)$

Consider i as a constant.

6. Originally Posted by Moo
Hello,

Substitute : $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i} \implies \sin(nx)=\frac{1}{2i} \cdot (e^{inx}-e^{-inx})$

$\int e^{ax} \sin(nx) dx=\frac{1}{2i} \cdot \left(\int e^{ax}e^{inx} dx-\int e^{ax}e^{-inx} dx\right)$ $=\frac{1}{2i} \cdot \left(\int e^{x(a+in)} dx-\int e^{x(a-in)} dx \right)$

Consider i as a constant.
I like that!

when i said complex analysis, i was thinking of using $e^{i \theta} = \cos \theta + i \sin \theta$

7. Actually $e^{ax}\sin(nx)=\text{Im}\,e^{(a+in)x}$ and Euler gets in the game.