$$If~\int\limits x^2cosx~dx = f(x) - \int\limits 2xsinx~dx$$then f(x) =
a) 2sinx + 2xcosx
b) x^2sinx
c) 2xcosx - x^2sinx
d) 4cosx - 2xsixs
e) (2-x^2)cosx - 4sinx

Question

$$If~\int\limits x^2cosx~dx = f(x) - \int\limits 2xsinx~dx$$then f(x) =
a) 2sinx + 2xcosx
b) x^2sinx
c) 2xcosx - x^2sinx
d) 4cosx - 2xsixs
e) (2-x^2)cosx - 4sinx

jigglypuff314 · Answer

.... >,<
I think it's B
from integrating the first part of it...
but I don't know why or how :/

ipwnbunnies · Answer

Hint: Do you recognize how the second part can be the second part of the integrations by parts formula? :3

vishweshshrimali5 · Answer

@jigglypuff314 Actually try figuring out the difference between integration by parts formula and the given integral.

jigglypuff314 · Answer

ah so it is B? :D
lel I was trying to do Tabular the first time >,<

jigglypuff314 · Answer

u = x^2            v = sinx
du = 2x dx     dv = cosx dx
uv - ∫v du
x^2sinx - ∫2xsinx dx

Thank you! :)

jigglypuff314 · Answer

O-O whyyy
integral of cosx = sinx
there would be no canceling negatives... ?

ipwnbunnies · Answer

Mhm, dv = cos x dx.
v = sin x + C
The minus sign is correct.

vishweshshrimali5 · Answer

Sorry it was correct.

vishweshshrimali5 · Answer

I misread the "v"

jigglypuff314 · Answer

hehehe xD I sometimes do that too :P
Thanks for your help <3