what is the integration of (xcos(x))
Have you heard of integration by parts?
\[\int vdu=uv-\int udv.\] lol
im a dissenter for no apparent reason; kinda like the jimmy dean kid :)
across, why did you delete yours ? now amistre looks foolish with that "lol" for no reason :)
ohh, i look foolish with or without it. its a trademark
down up cos(x) +x sin(x) = x sin(x) -1 -cos(x) = +cos(x) +0 -sin(x) = 0 int: x cos(x) dx = x sin(x) + cos(x) + C
to chk: d/dx: x sin(x) + cos(x) x' sin(x)+x sin(x)' + cos(x)' sin(x)+x cos(x) -sin(x) = x cos(x)
i think one day amistre will be an integrating god
To become an integrating god, one must first be able to find an anti-derivative of:\[\int e^{x^{2}}dx.\]:P
i think amistre can do it lol i can't so i will never reach goddess status
lol ..... i do that in my sleep, where reality and fantasy collide in an amalgamation of idiocy and genius
i felt like i didn't sleep last night i dreamed about programming i dreamed about having problems figuring out my code
That's bad, coz if you worked a lot on it - it's all gone when you wake up (even if you remembered to save your work often).
lol
i worked on it all last night and i don't remember what my brain concluded because my brain doesn't have a save button :(
@amistre one day i would like you to show me that up down thing. it is what you see in the movie "stand and deliver"
its what you do when you go ride the slides
its just my version of an IBP table: with multiple integration by parts; you tend to use the same u and v values such that u derives down to 0 and v just integrates up for the ride ....
\[\int udv=uv-(u'\int' v)+(u''\int'' v)-(u'''\int'''v)+(...\]if that makes any sense
yeah i have seen it but i guess i need to write it out for myself. if you ever saw the movie "stand and deliver" escalante does this on the board in like 6 seconds. crosses up and down
ill have to check it out :)
yeah, i tend to bump up the right side so i dont have to diagon it ...
which is why the first row is missing the "u"
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