How does integration by parts work?

[Tino]

??

would it be :

\frac{2}{9}sin(3x) + C

or

-\frac{2}{9}sin(3x) + C

 

[mattmns]

u = 2x
du = 2dx
dv = sin(3x)dx
v = -1/3cos(3x)

then you use uv - \int vdu

Which means you should have

(2x)(\frac{-1}{3}cos(3x)) - \int (\frac{-1}{3}cos(3x)(2dx))

That then equals

\frac {-2}{3}xcos(3x) + \frac{2}{3} \int (cos(3x)dx

So the last part would be: + \frac{2}{9}sin(3x) + C

Remember, you can pull out the constants for the integral. ie, -1, 1/3, and 2

 

[Tino]

Confused again !

Yesterday & Today in the post the answer was :

\frac{2}{9} sin(3x) - \frac{2}{3}x cos(3x) + C

And now the answer is:

-\frac {2}{3}xcos(3x) + \frac{2}{9}(sin(3x)dx

Which one is right

i know that they are the other way round from each other but which way can i put it ?

the question is :

\int2x\sin{(3x)}dx

and does it matter if i put it the wrong way round ?

 

[mattmns]

Both of those are correct.

Except of course the second one should have a + C

Tell me, what is the answer to these two problems:

10 - 8

-8 + 10

Both equal 2 would you agree?

 

[Tino]

Thanks

I get you now thanks very much !

 

[kkidd002]

help with integration by parts *urgent*

i need help with this because it's driving me crazy.
I need to integrate:

X^3(e^(3x^2))

I can do simpler integration by parts but i can't get this one to work out. The answer works out to:

[(x^2)/6 -1/18]e^(3x^2) please be very detailed, because I know how to do simpler integration by parts, but this example is proving to be very difficult.

 

[ace123]

Put this into a separate post