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.
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
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 before starting.
So it becomes which is easier to integrate.
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.