# Thread: Integration By Parts Problem

1. ## Integration By Parts Problem

Find the integral of (x cos(2x)) dx, using integration by parts.

I used x as dv/dx and cos(2x) as u.

So v = (v^2/2) and (du/dx) = -2sin(2x)

I am having problems using the formula:

I = uv - Integral ( u' v ) dx. Can anyone help me with this as I can't get the correct answer.

2. Originally Posted by haku
Find the integral of (x cos(2x)) dx, using integration by parts.

I used x as dv/dx and cos(2x) as u.

So v = (v^2/2) and (du/dx) = -2sin(2x)

I am having problems using the formula:

I = uv - Integral ( u' v ) dx. Can anyone help me with this as I can't get the correct answer.
why would you use x as dv??

let u = x and dv = cos(2x)

3. I thought it would be easier to integrate the x. I'll have a go with u=x, I'll post if I have any problems. Thanks.

4. Originally Posted by haku
I thought it would be easier to integrate the x. I'll have a go with u=x, I'll post if I have any problems. Thanks.
do not only consider what is easier to integrate, consider what you'll get when you are done. if you integrate x you will get x^2 times a trig function to integrate as opposed to just x times a trig function. you'd make your problem worst! if you differentiate x though, you will only have a trig function to integrate

5. Originally Posted by haku
I thought it would be easier to integrate the x. I'll have a go with u=x, I'll post if I have any problems. Thanks.
Yes, it's easier to integrate it but it won't bring you anywhere. But if we differentiate x, it'll be 1 and if we differentiate 1, we get 0. This means you will surely get a result using integration by parts 2 times. Learn tabular integration, it makes easier to use multiple partial integrations.

If you have troubles integrating this, you can substitute $\displaystyle t=2x$ before starting.

$\displaystyle \int x \cos 2x ~dx$
$\displaystyle t=2x$
$\displaystyle dt = 2dx$
$\displaystyle x=\frac{t}{2}$

So it becomes $\displaystyle \frac{1}{4} \int t \cos t ~dt$ which is easier to integrate.

6. Originally Posted by wingless
Yes, it's easier to integrate it but it won't bring you anywhere. But if we differentiate x, it'll be 1 and if we differentiate 1, we get 0. This means you will surely get a result using integration by parts 2 times. Learn tabular integration, it makes easier to use multiple partial integrations.

If you have troubles integrating this, you can substitute $\displaystyle u=2x$ before starting.

$\displaystyle \int x \cos 2x ~dx$
$\displaystyle u=2x$
$\displaystyle du = 2dx$
$\displaystyle x=\frac{u}{2}$

So it becomes $\displaystyle \frac{1}{4} \int u \cos u ~du$ which is easier to integrate.
you should use t or something for the substitution, since the poster uses u for the integration by parts formula. there may be confusion.

7. Originally Posted by Jhevon
you should use t or something for the substitution, since the poster uses u for the integration by parts formula. there may be confusion.
Haha, you're right. Edited

8. So for this example:

Integral (Limits 0 to 1) (2x-1) e^(3x+2).

I set (dv/dx)=(2x-1) so, v=(((2x-1)^2)/2)

and u=e^(3x+2) so, u'=3e^(3x+2).

Am I correct so far. Then using the formula:

e^(3x+2).(((2x-1)^2)/2) - (Integral of) (3e^(3x+2)).(((2x-1)^2)/2).

Would you use integration by parts again for the second part? I can't seem to do this. Please help.

9. $\displaystyle \int^{1}_{0} (2x-1)e^{3x+2}~dx$

First simplify the function. This will make it really easy to integrate.

$\displaystyle \int^{1}_{0} (2x-1)e^{3x}e^{2}~dx$

$\displaystyle e^{2}\int^{1}_{0} (2x-1)e^{3x}~dx$

Let $\displaystyle u=2x-1$, $\displaystyle dv = e^{3x}$ and integrate it by parts two times.

10. First, keep simplifying.

$\displaystyle \int{(2x-1)e^{3x}}\;dx\;=\;2\int{xe^{3x}}\;dx\;-\;\int{e^{3x}}\;dx\;=\;$

Second, why twice?

$\displaystyle \int{xe^{3x}}\;dx\;=\;\int{x}\;d\left(\frac{e^{3x} }{3}\right)\;=\;\frac{xe^{3x}}{3}\;-\;\int{\frac{e^{3x}}{3}}\;dx$