xsin x +cos x +C
Why is it that those that get the anti-derivative right also forget the constant?????
2006-07-15 02:54:33
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answer #1
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answered by mathematician 7
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Integrate Xcosx
2016-11-11 01:42:56
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answer #2
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answered by dierks 4
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Integral Of X Cos X
2017-01-01 08:07:30
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answer #3
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answered by ? 4
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This Site Might Help You.
RE:
integration of xcosx?
integration
2015-08-14 09:45:30
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answer #4
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answered by Tucky 1
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Integration by parts. The function is a product of two functions:
(x) * (cosx)
Call one of them "u" and the other "dv"
Choose the "u" so that its derivative simplifies
u = x
du = 1*dx
dv = cosx
v = sinx
This is for the formula setup, which is
u*v - Integral (vdu)
=xsinx - Integral(sinx * 1 * dx)
=xsinx - (-cosx) + C
= xsinx + cosx + C
2006-07-15 12:23:53
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answer #5
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answered by Anonymous
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⌠ x cosx dx
Use the intergration by parts formula: ⌠u dv = uv - ⌠v du
Let u = x
du/dx = 1 → du = dx
Let dv = cosx dx
⌠ dv = ⌠ cos x dx
v = sin x
Now fill in the information into the formula:
⌠u dv = uv - ⌠v du
⌠ x cosx dx = x sinx - ⌠ sinx dx
= x sinx - (-) cosx
= x sinx + cosx (Answer)
Now differentiate the answer with the following formula:
uv = u dv/dx + v du/dx
Let u = x and v = Sin x
du/dx = x cosx + sinx(1) + (- sinx)
= x cosx (Answer confirmed).
2006-07-15 08:58:13
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answer #6
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answered by Brenmore 5
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Try integration by "parts".
2006-07-15 01:05:46
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answer #7
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answered by bobweb 7
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int udv = uv- int vdu
intgral of cox = sin x
cosx = d(sinx)/dx
int xcosx dx = int x d(sinx) = x.sinx - int(sinx.1. dx)
= x sinx - int(sinxdx)
= xsinx + cos x(integral of sin x = -cos x) +c where c is arbritary constant
2006-07-15 01:11:54
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answer #8
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answered by Mein Hoon Na 7
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xsinx+cosx
2006-07-15 01:45:09
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answer #9
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answered by omar g 2
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I guess,
x^2/2 * sin x
2006-07-15 01:21:18
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answer #10
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answered by nayanmange 4
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ans: x sinx + c ( reason for c is because this is an indefinite integral)
2006-07-15 02:05:34
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answer #11
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answered by -mystic- 2
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