# Integration by parts question

1. Jun 3, 2010

### stripes

1. The problem statement, all variables and given/known data

Gosh I've been asking a lot of questions lately...anyways...

I tried this question two separate times and couldn't manage to figure out where i went wrong...

$$\int e^{-x}cos2x dx$$

2. Relevant equations

$$uv - \int v du = \int u dvdx$$

3. The attempt at a solution

let $u = e^{-x}$ and $dv = cos2x dx$
so $du = -e^{-x}$ and $v = \frac{1}{2}sin2x$

$$\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x + \frac{1}{2}\int e^{-x}sin2x dx$$

use integration by parts again, this time let $u = e^{-x}$ and $dv = sin2x dx$ so $du = -e^{-x}$ and $v = \frac{-1}{2}cos2x$

so now we have:

$$\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x - \frac{1}{2}\int e^{-x}cos2x dx$$

bring the $\frac{1}{2}\int e^{-x}cos2x dx$ to the LHS and solve for the integral.

$$\frac{3}{2}\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x$$

so finally:

$$\int e^{-x}cos2x dx = \frac{1}{3}e^{-x}(sin2x - cos2x)$$

which is incorrect... at which step did I go wrong?

Last edited: Jun 3, 2010
2. Jun 3, 2010

### Cyosis

You forgot a factor 1/2 the second time you used integration by parts.

3. Jun 3, 2010

### stripes

ahh, in which case i would get

$$\int e^{-x}cos2x dx = \frac{1}{5}e^{-x}(2sin2x - cos2x)$$

4. Jun 3, 2010

### stripes

thanks for pointing out my silly mistake...I have one more question that i will post up in a few minutes, this assignment is really killing me. Thanks again cyosis.

5. Jun 3, 2010

### Staff: Mentor

Plus the constant of integration.