# Integration by parts problem

1. Nov 30, 2006

### lord12

integral of (e^(2x))(cos3x)dx

I get 1/2e^(2x)sin2x - 3/2integral(e^(2x)cos3xdx)

what do i do next?

2. Nov 30, 2006

Let $$u = e^{2x}$$ $$dv = \cos 3x$$ $$du = 2e^{2x}$$ $$v = \frac{1}{3}\sin 3x$$

Since $$\int udv = uv-\int vdu$$, we have:

$$\frac{e^{2x}}{3}\sin 3x - \frac{2}{3}\int e^{2x}\sin 3x = \int e^{2x}\cos 3x \; dx$$

Now apply integration by parts to $$\frac{2}{3}\int e^{2x}\sin 3x$$ and you can solve for the integral.

Last edited: Nov 30, 2006
3. Nov 30, 2006

### arildno

Please be more careful with what you write!
$$I=\int{e^{2x}\cos(3x)dx=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}\int{e}^{2x}\sin(3x)dx+C$$
where C is some integration constant.

Now, the right-hand side indefinite integral may be rewritten as:
$$\int{e}^{2x}\sin(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}\int{e}^{2x}\cos(3x)dx=\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}I$$
And your first expression may then be written as:

$$I=\frac{1}{2}e^{2x}\cos(3x)+\frac{3}{2}(\frac{1}{2}e^{2x}\sin(3x)-\frac{3}{2}I)+C$$

Now, solve for I.

4. Dec 4, 2006

### Gib Z

I think lord12s main problem was that he has never encountered a situation where he has to apply integration by parts twice. It occurs often, and many times more than just once. In fact many go on an infinite number of applications.

5. Dec 4, 2006

It would definitely be interesting to see what a man who applied integration by parts for an infinite number of times would look like after he did so. :tongue2:

6. Dec 5, 2006

### Gib Z

lol yea that would be. But in case your looking for one where integration is applied $n$ times, and $n$ can be as large as you want, then look at Euler's ...Totient Function? I can't remember the name, but it basically stated:

$$\int e^{-x} x^n = n!$$

I meant to write, when integrated from infinity to 0, but dont know how..

anyway its proven by applying Integration by parts n times, try it. Btw this leads us to be able to calculate factorials all positive real numbers, not just integers. Its very tedious to do approximate by hand, but in case you want to, just to impress someone that you can find the factorial of 5.6, then use the trapezoidal rule, much simpler and accurate than simpson's rule in this case. And approximate by finding area of 0 to 4n, infinity may be a bit hard...I dont know why the function seems to rapidly dive down to zero at 4n, maybe someone can answer that for me?

Last edited: Dec 5, 2006