# Integrate

• Sep 1st 2009, 07:47 PM
arze
Integrate
$\int e^x \sin 2x dx$
I used the integration by parts method.
$\int v\frac{du}{dx}=(uv)-\int u\frac{dv}{dx}$
I let $v=\sin 2x\rightarrow \frac{dv}{dx}=2\cos 2x$
and $\frac{du}{dx}=e^x\rightarrow u=e^x$
so,
$uv-\int u\frac{dv}{dx}$
$=e^x(\sin 2x)-\int 2e^x\cos 2x dx$
This is the part I am stuck at. If I integrate $2e^x\cos 2x$ by parts I get another product. And it will continue for I don't know what because I can't get rid of the $e^x$.
Any help? I've definitely done something wrong. Thanks
• Sep 1st 2009, 07:50 PM
Random Variable
You'll get the same integral on both sides of the equation.
• Sep 1st 2009, 07:51 PM
pomp
Quote:

Originally Posted by arze
$\int e^x \sin 2x dx$
I used the integration by parts method.
$\int v\frac{du}{dx}=(uv)-\int u\frac{dv}{dx}$
I let $v=\sin 2x\rightarrow \frac{dv}{dx}=2\cos 2x$
and $\frac{du}{dx}=e^x\rightarrow u=e^x$
so,
$uv-\int u\frac{dv}{dx}$
$=e^x(\sin 2x)-\int 2e^x\cos 2x dx$
This is the part I am stuck at. If I integrate $2e^x\cos 2x$ by parts I get another product. And it will continue for I don't know what because I can't get rid of the $e^x$.
Any help? I've definitely done something wrong. Thanks

Call the original integral (the one you want to calculate) I, and then proceed as you did. Now perform integration by parts on what you arrived at, and you will end up with an integral which is a scalar multiple of the original integral you called I. Rearrange and you will arrive at your answer.

This is a well known trick for dealing with these, and other similar types of integrals.

Good luck.
• Sep 1st 2009, 07:57 PM
arze
Quote:

Originally Posted by pomp
Call the original integral (the one you want to calculate) I, and then proceed as you did. Now perform integration by parts on what you arrived at, and you will end up with an integral which is a scalar multiple of the original integral you called I. Rearrange and you will arrive at your answer.

This is a well known trick for dealing with these, and other similar types of integrals.

Good luck.

Thanks! But I think i haven't learned this trick.
so
$I=e^x(\sin 2x)-\int 2e^x\cos 2x dx$
then I integrate $2e^x\cos 2x dx$?
Thanks
• Sep 1st 2009, 08:03 PM
pomp
Quote:

Originally Posted by arze
Thanks! But I think i haven't learned this trick.
so
$I=e^x(\sin 2x)-\int 2e^x\cos 2x dx$
then I integrate $2e^x\cos 2x dx$?
Thanks

Call the original integral I,

$I = \int e^x \sin{2x} dx$
• Sep 1st 2009, 08:18 PM
arze
Sorry I don't understand. From my workings, $I=e^x(\sin 2x)-\int 2e^x\cos 2x dx$? Then what's next?
• Sep 1st 2009, 08:25 PM
Random Variable
$\int e^{x} \sin 2x \ dx = e^x \sin 2x - 2 \int e^{x} \cos 2x \ dx = e^x \sin 2x -2e^{x} \cos 2x - 4 \int e^{x} \sin 2x \ dx$

then $5 \int e^{x} \sin 2x \ dx = e^x \sin 2x -2e^{x} \cos 2x$

so $\int e^{x} \sin 2x \ dx = \frac{1}{5}e^{x} \sin 2x - \frac{2}{5} e^{x} \cos 2x + C$