How do you integrate e^x sin 2x using a special case?

[BOAS]

Hello,

I have been going round in circles on a problem, but I am aware of a special case regarding this integral, however I don't know how to use or really understand it.

Homework Statement

Integrate the following function with respect to x.

$e^{x} \sin 2x$

Homework Equations

$2 \int e^{x} \sin x dx = e^{x} (\sin x - \cos x) + c$

I believe this is the relevant equation I need to use...

The Attempt at a Solution

$\int e^{x} \sin 2x = uv - \int u \frac{dv}{dx} dx$

$v = e^{x}$

$\frac{dv}{dx} = e^{x}$

$\frac{du}{dx} = \sin 2x$

$u = -\frac{1}{2} \cos 2x$

So,

$\int e^{x} \sin 2x = -\frac{1}{2} e^{x} \cos 2x - \int -\frac{1}{2} e^{x} \cos 2x$

Consider

$\int -\frac{1}{2} e^{x} \cos 2x = uv - \int u \frac{dv}{dx} dx$

$v = e^{x}$

$\frac{dv}{dx} = e^{x}$

$\frac{du}{dx} = -\frac{1}{2} \cos 2x$

$u = -\frac{1}{4} \sin 2x$

so,

$\int -\frac{1}{2} e^{x} \cos 2x = -\frac{1}{4} e^{x} \sin 2x - \int -\frac{1}{4} e^{x} \sin 2x$

Bringing that together gives,

$\int e^{x} \sin 2x = -\frac{1}{2} e^{x} \cos 2x + \frac{1}{4} e^{x} \sin 2x - \int -\frac{1}{4} e^{x} \sin 2x$


This doesn't seem to be getting me anywhere as I'm sure repeating this process will leave me with another integral on the end...

I was trying to follow the process my book uses to arrive at the 'special case' I gave in the relevant equations section, but I don't know how to proceed.

Thanks!

 

[Office_Shredder]

$\int e^{x} \sin 2x = -\frac{1}{2} e^{x} \cos 2x + \frac{1}{4} e^{x} \sin 2x - \int -\frac{1}{4} e^{x} \sin 2x$

Subtracting$\frac{1}{4} \int e^{x} \sin 2x dx$ to both sides

$$\frac{3}{4} \int e^{x} \sin 2x dx = -\frac{1}{2} e^{x} \cos 2x + \frac{1}{4} e^{x} \sin 2x+C$$

However I think you have a minus sign issue in getting to that last equation you should go back and look for. The final step will be exactly the same idea though.