Where does the extra factor of 3 come from in this e^(2x)cos(2x) integral?

[TSN79]

I'm trying to perform the following integral
$$\pi \int\limits_0^\pi {e^{2x} } \left( {\frac{1}{2} - \frac{1}{2}\cos 2x} \right)dx$$
I split the integral and temporarely ignore the Pi so that I get
$$\frac{1}{2}\int {e^{2x} dx} - \frac{1}{2}\int {e^{2x} \cdot \cos } \left( {2x} \right)dx$$
Now, using partial integration on the second part I get
$$\int {e^{2x} \cdot \cos \left( {2x} \right)} dx = \frac{1}{2}e^{2x} \cdot \sin \left( {2x} \right) - \int {\sin \left( {2x} \right)} \cdot e^{2x} dx$$
Using partial integration again on the right integral I get
$$\int {\sin \left( {2x} \right)} \cdot e^{2x} dx = - \frac{1}{2}e^{2x} \cdot \cos \left( {2x} \right) + \int {\cos \left( {2x} \right) \cdot e^{2x} dx}$$
I appears I haven't gotten anywhere, but I can now combine the last two lines and get
$$\begin{array}{l}<br /> \int {\cos \left( {2x} \right) \cdot e^{2x} dx = \frac{1}{2}e^{2x} \cdot \sin \left( {2x} \right) - (\frac{1}{2}e^{2x} \cdot \cos \left( {2x} \right)} - \int {\cos \left( {2x} \right) \cdot e^{2x} \left. {dx} \right)} \\ <br /> 2\int {\cos \left( {2x} \right) \cdot e^{2x} dx = \frac{1}{2}e^{2x} } \cdot \sin \left( {2x} \right) - \frac{1}{2}e^{2x} \cdot \cos \left( {2x} \right) \\ <br /> \int {\cos \left( {2x} \right)} \cdot e^{2x} dx = \frac{1}{4}e^{2x} \cdot \sin \left( {2x} \right) - \frac{1}{2}e^{2x} \cdot \cos \left( {2x} \right) \\ <br /> \end{array}$$
Finally, multiplying in the Pi and the first initial half of the integral:
$$\pi \cdot \left( {\frac{1}{4}e^{2x} - \frac{1}{2}\left( {\frac{1}{4}e^{2x} \cdot \sin \left( {2x} \right) - \frac{1}{4}e^{2x} \cdot \cos \left( {2x} \right)} \right)} \right)$$
Putting in Pi and 0 for x, and subtracting the two, I arrive at this expression:
$$\frac{{3\pi \left( {e^{2\pi } - 1} \right)}}{8}$$
The problem is that this factor 3 shouldn't be there. If you just perform the initial integration on a calculator the answer is the same except for the factor 3, so where am I going wrong here?

 

[ghotra]

You messed up a couple minus signs. I'm going to copy and paste your code with corrections.
$$\begin{array}{l}<br /> \int {\cos \left( {2x} \right) \cdot e^{2x} dx = \frac{1}{2}e^{2x} \cdot \sin \left( {2x} \right) - (-\frac{1}{2}e^{2x} \cdot \cos \left( {2x} \right)} + \int {\cos \left( {2x} \right) \cdot e^{2x} \left. {dx} \right)} \\ <br /> 2\int {\cos \left( {2x} \right) \cdot e^{2x} dx = \frac{1}{2}e^{2x} } \cdot \sin \left( {2x} \right) + \frac{1}{2}e^{2x} \cdot \cos \left( {2x} \right) \\ <br /> \int {\cos \left( {2x} \right)} \cdot e^{2x} dx = \frac{1}{4}e^{2x} \cdot \sin \left( {2x} \right) + \frac{1}{4}e^{2x} \cdot \cos \left( {2x} \right) \\ <br /> \end{array}$$

That should point you in right direction. Eventually, you should have something like this:

$$\left.\frac{1}{4} \mathrm{e}^{2x} \right|_0^\pi - \left.\frac{1}{8}\mathrm{e}^{2x}\left(\cos 2x + \sin 2x\right)\right|_0^\pi$$

So, you'll get something: 2Y - Y = Y where Y is the answer you expect.