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p3 integration

Use integration by parts to find xcos2x dx

Prove that the answer may be expressed as 1/2sinx(2xcosx - sinx) + c

Please can someone provide the solution

Also does anyone have the markscheme for the May 2002 P3 Edexcel paper?
If they do please PM me
Reply 1
Avatar for El Stevo
El Stevo
11
Int xcos(2x) dx

Int uv' = uv - Int u'v dx

u = x
u' = 1
v' = cos(2x)
v = (1/2)sin(2x)

Int xcos(2x) dx = (1/2)xsin(2x) - Int (1/2)sin(2x) dx
Int xcos(2x) dx = (1/2)xsin(2x) - (-(1/4)cos(2x) + c
Int xcos(2x) dx = (1/2)xsin(2x) + (1/4)cos(2x) + c

using double angle formulae...

(1/2)xsin(2x) + (1/4)cos(2x) + c = xsinxcosx + (1/2){(cos^2)x - (sin^2)x + c}
= xsinxcosx + 1/2 - (1/2)(sin^2)x - (1/2)(sin^2)x + c
= xsinxcosx - (1/2)(sin^2)x + c + 1/2
= xsinxcosx - (1/2)(sin^2)x + k
= (1/2)sinx(2xcosx - sinx) + k
Reply 2
Avatar for Aitch
Aitch
2
El Stevo
Int xcos(2x) dx

Int uv' = uv - Int u'v dx

u = x
u' = 1
v' = cos(2x)
v = (1/2)sin(2x)

Int xcos(2x) dx = (1/2)xsin(2x) - Int (1/2)sin(2x) dx
Int xcos(2x) dx = (1/2)xsin(2x) - (-(1/4)cos(2x) + c
Int xcos(2x) dx = (1/2)xsin(2x) + (1/4)cos(2x) + c

edit: thinking whether you meant x(cos^2)x or how to fiddle this answer...


I get ½ * sinx(2xcosx - sinx) + c (where c = k+¼)

Aitch

I'd type it up, but someone else will..
Reply 3
Avatar for El Stevo
El Stevo
11
Aitch
I get ½ * sinx(2xcosx - sinx) + c (where c = k+¼)

Aitch

I'd type it up, but someone else will..


as did i :smile:, now i typed it up...
Reply 4
Avatar for Aitch
Aitch
2
Aitch
I get ½ * sinx(2xcosx - sinx) + c (where c = k+¼)

Aitch

I'd type it up, but someone else will..


If noone else does it in the next 10 minutes, I'll type it up... Should be doing P5 qs...

Aitch
Reply 5
Avatar for Aitch
Aitch
2
El Stevo
as did i :smile:, now i typed it up...


Thanks! Now I will go and do those P5 qs!

Aitch
Reply 6
Avatar for Fermat
Fermat
8
Int xcos(2x) dx = (1/2)xsin(2x) + (1/4)cos(2x) + c

put sin(2x) = 2sinx.cosx
and cos(2x) = 1 - 2sin²x

then work it out

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