[P3] Integration - by parts? Watch

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MdSalih
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#1
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#1
I have an equation:

y = 2x^(1/2).Sin(x)

Need to integrate it. Not fully sure how to though. If i attempt to do it by parts, it seems to me like it will go around in circles.

Anyone care to shed some light on it? Thanks.

MdSalih
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pkchips
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#2
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that does seem rather peculiar.

In what context did u find this question? Are you sure you aren't meant to be doing integral of y² (volume of revolution) ?

if not, i don't know how you would do it.
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MdSalih
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nup.. not a volume :-/ - It's just a straight integral...

MdSalih
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Nima
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(Original post by MdSalih)
I have an equation:

y = 2x^(1/2).Sin(x)
Let u = 2x^(1/2) -> du/dx = (-1/2)(2)x^(-1/2) = -x^(-1/2)
Let dv/dx = sinx -> v = -cosx

Int. y dx. = -2x^(1/2).cosx - Int. x^(-1/2).cosx. dx.

To find Int. x^(-1/2).cosx dx.
Let u = cosx -> du/dx = -sinx
Let dv/dx = x^(-1/2) -> v = 2x^(1/2)

-> Int. x^(-1/2).cosx dx. = 2x^(1/2)cosx + 2 Int. x^(-1/2)sinx dx. + c

Hence: Int. y dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx. + d
-> 2 Int. x^(1/2).sinx dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx.
-> 4 Int. x^(1/2).sinx dx. = -4x^(1/2)cosx + d

-> Int. x^(1/2).sinx dx = k - x^(1/2).cosx
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pkchips
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(Original post by Nima)
Let u = 2x^(1/2) -> du/dx = (-1/2)(2)x^(-1/2) = -x^(-1/2)
Although I haven't read the rest of your solution (although it may well be right otherwise), when you differentiate 2x^0.5, im pretty sure u get x^(-0.5) and not -x^(-0.5).
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Gaz031
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(Original post by Nima)
Let u = 2x^(1/2) -> du/dx = (-1/2)(2)x^(-1/2) = -x^(-1/2)
Let dv/dx = sinx -> v = -cosx

Int. y dx. = -2x^(1/2).cosx - Int. x^(-1/2).cosx. dx.

To find Int. x^(-1/2).cosx dx.
Let u = cosx -> du/dx = -sinx
Let dv/dx = x^(-1/2) -> v = 2x^(1/2)

-> Int. x^(-1/2).cosx dx. = 2x^(1/2)cosx + 2 Int. x^(-1/2)sinx dx. + c

Hence: Int. y dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx. + d
-> 2 Int. x^(1/2).sinx dx. = -2x^(1/2).cosx - 2x^(1/2)cosx - 2 Int. x^(-1/2)sinx dx.
-> 4 Int. x^(1/2).sinx dx. = -4x^(1/2)cosx + d

-> Int. x^(1/2).sinx dx = k - x^(1/2).cosx
As the above poster said, there was a slip when differentiating 2x^0.5
If you were to do it again with the error corrected you'd find that when you do your second integration by parts you're simply reversing the first one and your string of function on the right would cancel out, so you won't have anything on the right.
Can i ask where the question is from?
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smartguyhirz
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yup right substitution, jus the diff. error otherwise all good.
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Aitch
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Web Mathematica baulks at it.

see

http://www.calc101.com/webMathematica/integrals.jsp

...which is usually ominous.

Aitch
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El Stevo
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Int 2x^(1/2).Sin(x) dx
= 2 Int [x^(1/2)]sinx dx
==========================
Int [x^(1/2)]sinx dx

t = sqrtx => x = (t^2) => dx = 2t dt

Int [x^(1/2)]sinx dx
= Int t.sin(t^2).2t dt
= Int 2(t^2).sin(t^2) dt

ergo...

Int 2x^(1/2).Sin(x) dx
= 2 Int 2(t^2).sin(t^2) dt
= Int 4(t^2).sin(t^2) dt

u' = 4sin(t^2)
u = -(2/t])cos(t^2)
v = (t^2)
v' = 2t

Int (t^2).4sin(t^2) dt
= [(t^2)][-(2/t])cos(t^2)] - Int [2t][-(2/t])cos(t^2)]
= -(2t)cos(t^2) - Int [-4cos(t^2)]
= -(2t)cos(t^2) + Int 4cos(t^2)
= -(2t)cos(t^2) + (2/t)sin(t^2)
= (2/t)sin(t^2) - (2t)cos(t^2)

t = sqrtx, ergo:

(2/t)sin(t^2) - (2t)cos(t^2)) = 2(x^[-1/2])sinx - 2(x^[1/2])cosx

Int 2x^(1/2).Sin(x) dx = 2(x^[-1/2])sinx - 2(x^[1/2])cosx + C
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Gaz031
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u' = 4sin(t^2)
u = -(2/t)cos(t^2)
There's a problem here. t isn't a constant so you can't do that.
If you were to differentiate 'u' you would have to use the product rule and would get something different to u'.
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El Stevo
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(Original post by Gaz031)
There's a problem here. t isn't a constant so you can't do that.
If you were to differentiate 'u' you would have to use the product rule and would get something different to u'.
doh! - i just worked it out...
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!Laxy!
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#12
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Is it not just:

4/3x^3/2.sinx + C
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idiopathic
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(Original post by !Laxy!)
Is it not just:

4/3x^3/2.sinx + C
Unfortunately not.
Differentiate it, using the product rule, and you'll see it's not the same as the original integral.
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