please derive this integral.. thanks a lot!

**Chop Suey** · August 25th 2008, 08:07 AM

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.

**vicky0514** · August 25th 2008, 08:11 AM

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!

**Jhevon** · August 25th 2008, 08:21 AM

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

**Moo** · August 25th 2008, 10:24 AM

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.

**Jhevon** · August 25th 2008, 10:30 AM

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 $e^{i \theta} = \cos \theta + i \sin \theta$

**Krizalid** · August 25th 2008, 10:49 AM

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

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