sarax
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#1
Report Thread starter 13 years ago
#1
ok here goes, I have a problem that I just can't do, can anyone help me out

∫x2sin(1/2x) dx

but between the boundries 2pi and pi (i don't know how to show that on a comp)

can anyone please help
sara
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Knogle
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#2
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Is that \Huge\int_\pi^{2\pi} x^2 \sin\left(\frac{1}{2x}\right) dx?
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sarax
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#3
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x/2 instead of 1/2x but the rest is right yeah
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meef cheese
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u = x => du/dx = 1

dv/dx = sin(x/2) => v = -2cos(x/2)

I = uv - u'v
= -2xcos(x/2) + 2 INT cos(x/2) dx
= -2xcos(x/2) + 4 sin(x/2)

and then plug in the limits to get 4pi - 4
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sarax
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#5
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i see your point, but doesn't u need to equal x^2 not x....
and then it doesn't work...????
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meef cheese
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Ahh, sorry I didn't see that it was x2.

Let I denote the integral and integrate by parts twice. [ie. once to get it into terms of cos(x/2), then again to get it back into terms of sin(x/2)]. You'll have I = ... on the left hand side and some fraction of I on the RHS. Bring them over to the same side and that shold give you a solution.
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sarax
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#7
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ok so i just have to do it twice...
if i stick my answer up will ya tell me if its right??
sara
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sarax
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#8
Report Thread starter 13 years ago
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(Original post by e-unit)
u = x => du/dx = 1

dv/dx = sin(x/2) => v = -2cos(x/2)

I = uv - u'v
= -2xcos(x/2) + 2 INT cos(x/2) dx
= -2xcos(x/2) + 4 sin(x/2)

and then plug in the limits to get 4pi - 4
shouldn't it be - not +

ok i did it for teh first time and got

-2x^2 cos(x/2) - INT -2cos(x/2)2x

can I put the two in front of the integral sign??
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meef cheese
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#9
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(Original post by sarax)
shouldn't it be - not +

ok i did it for teh first time and got

-2x^2 cos(x/2) - INT -2cos(x/2)2x

can I put the two in front of the integral sign??
Yes and the two negatives cancel.
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sarax
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#10
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i ended up with

-2x62 cos(x/2) -8x sin (x/2) + 16 sin(x/2)
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sarax
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#11
Report Thread starter 13 years ago
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actually no
i got this
-2x^2 cos(x/2) + 8x sin(x/2) -16cos(x/2)
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