You are not logged in.
Pages: 1
I am guessing integration by parts, but I am stuck.
Last edited by Gotovina (2008-03-10 10:52:25)
Offline
Integration by parts it is.
Choose x to be the differentiated bit and cos(2x) to be the integrated bit.
That way x disappears and you're left with an integral that can be evaluated directly.
Why did the vector cross the road?
It wanted to be normal.
Offline
Yes by parts.
The point of integration by parts is letting u equal a term that when you integrate it, it becomes simpler, in this case x.
so by integration by parts this is equal to:
Last edited by LuisRodg (2008-03-11 13:33:55)
Offline
LaTeX was down temporarily due to the server switch, but it should work fine now. Test:
Why did the vector cross the road?
It wanted to be normal.
Offline
Mathsy, could you tell me whats wrong with my code?
Offline
You could also integrate by U-sub. if you were like me and in Calc. AB and didn't know parts yet.
But if that were the case you would have to sub U back in a second time.
Good luck.
(>^^)> here's a kirby for you.
e...the red-headed stepchild of math.
Offline
Mathsy, could you tell me whats wrong with my code?
Sorry, I should have realised to check that.
I think the problem is to do with line breaks. LaTeX needs backslashes at the end of lines or it goes funny.
You have:
u = x
dv = \cos(2x)dx
du = dx
v = \frac{1}{2}\sin(2x)
, which gives
.Instead, do:
u = x \\
dv = \cos(2x)dx\\
\\
du = dx\\
v = \frac{1}{2}\sin(2x)
, to get
Why did the vector cross the road?
It wanted to be normal.
Offline
Ok I fixed the first part and tried to do the same to the second but still showing up messed up?
Offline
You needed to put a space between \int and vdu. Also, if you don't want an indent in the first line like above, put \\ at the beginning of the line.
Last edited by Daniel123 (2008-03-11 07:43:08)
Offline
Thanks Daniel123.
So I guess Gotovina got her answer and more than she asked for....:)
Offline
Pages: 1