The Student Room Group

integration by parts

3xsin2xdx\int3xsin2xdx with limits 1/2 pi and zero

[-3/2cos2x+3/4sin2x] is what i end up as my integral then when i sub the limits i keep getting a three for my answer. Can't get it to latex properly to show my working, but does that integral seem correct?
Looks right.
Original post by Brit_Miller
Looks right.


am i just subbing in the limits wrong or something stupid like that; what do you get when you sub the limits into that?
Original post by Emissionspectra
am i just subbing in the limits wrong or something stupid like that; what do you get when you sub the limits into that?


I'll write it out, it just looked right looking at it, I'd better check. :biggrin:
Reply 4
Original post by Emissionspectra
3xsin2xdx\int3xsin2xdx with limits 1/2 pi and zero

[-3/2cos2x+3/4sin2x] is what i end up as my integral then when i sub the limits i keep getting a three for my answer. Can't get it to latex properly to show my working, but does that integral seem correct?


I think ur missing an x here. :smile: -(3/2)xcos(2x)
Reply 5
Original post by Emissionspectra
3xsin2xdx\int3xsin2xdx with limits 1/2 pi and zero

[-3/2cos2x+3/4sin2x] is what i end up as my integral then when i sub the limits i keep getting a three for my answer. Can't get it to latex properly to show my working, but does that integral seem correct?


Original post by Brit_Miller
Looks right.


Mistake here.

It should be, [3xcos2x2]0π2+[34sin2x]0π2[br][br][br] \displaystyle \left[\frac{-3xcos2x}{2}\right] ^{\frac{\pi}{2}}_0 + \left[\frac34sin2x\right]^{\frac{\pi}{2}}_0[br][br][br]
Original post by SilentIdea
I think ur missing an x here. :smile:


Yep, just realised that when I wrote it out.
Original post by SilentIdea
I think ur missing an x here. :smile: -(3/2)xcos(2x)


Ye i see it now urgh ah well got it correct now; thanks :smile:

Quick Reply

Latest