Quick question on integration by parts

[t_n_p]

Homework Statement

I'm following an example in the textbook that states:

http://img24.imageshack.us/img24/1672/33686252.jpg

I was just wondering what happened to the 2 out the front, I would have been more inclined to think this would be the next step:

http://img34.imageshack.us/img34/4854/a1aa.jpg

Any explanations?

 

[lanedance]

when you perform the integration by parts you get the original integral on both sides of the equation, which leads to the cancelling of the factor of 2 on the right hand side

 

[t_n_p]

I'm not following 100%.
The integral on the RHS is not the original integral on the LHS?

This it the formula I used (and I am familar with):

(of course with respect to t here not x)

fyi: I let f = Gsintcost and g' = Gdot

everything works out fine, its just the 2 that has me puzzled

 

[lanedance]

ok so in some bastardised short hand

du = G'
u = G
v = Gsc
dv = G'sc + G(c² - s²)

\int du.v = uv| - \int u.dv

\int G'.Gsc = G.Gsc| - \int G.(G'sc + G(c^2-s^2))

\int G'.Gsc = G^2 sc| - \int G.G'sc -\int G^2(c^2-s^2)

2 \int G'.Gsc = G^2 sc| - \int G^2(c^2-s^2)

2 \int G'.Gsc = G^2 sc| + \int G^2(s^2-c^2)

 

[t_n_p]

I think dv should be dv = G'sc + G(c^2 - s^2)

 

[lanedance]

yeah i think you're right, but hopefully the bit about the orginal intergal aappearing on the RHS is celar

 

[t_n_p]

yeah i think you're right, but hopefully the bit about the orginal intergal aappearing on the RHS is celar

bugger, I think dv = G'sc + G(c^2 - s^2), but then if I change that, then I don't get the correct answer!

I understand what you mean by the original integral showing up on the RHS (cos now I can take it over to the lhs and get 2* original integral).


edit: found the mistake. When I broke up the integral I forgot to take the minus inside both.

got it all sorted now.

Cheers!

 

[lanedance]

no worries - i updated the working above as well