Integration by Parts

  • Thread starter adamwitt
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  • #1
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Homework Statement



Find the solution to:

y' = x.y.cos(x^2)


Homework Equations



Integration by Parts method.


The Attempt at a Solution



Step 1
(dy/dx).(1/y) = x.cos(x2)

(1/y) dy = x.cos(x2) dx

Step 2
Integrate both sides.

ln|y| = integratal of [ x.cos(x2) dx ]

Step 3
Using integration by parts...

u = cos(x2) => du = -2x.sin(2^x)
dv = x => v = 1.dx

Step 4
Subbing back in...

ln|y| = u.v - integral of v.du = cos(x2) - integral of [-2x.sin(x2)]

ln(y) = cos(x2) + 2*integral of [x.sin(x2)]

Step 5
Using integration by parts a second time...

u = sin(x2) => du = 2x.cos(x2)
dv = x => v = 1.dx

Step 6
Subbing back in...

ln|y| = cos(x2) + 2 ( sin(x2) - 2*integral of x.cos(x2).dx )


Step 7

I stop my attempt there because it just seems to eventually I get to a point where it becomes a function of itself? (ie. the x.cos(x2)

What do I do next? or have I gone wrong somewhere?


EDIT --- SOME FURTHER WORK, IS THIS CORRECT?

Ok so I noticed that we get "integral of [x.cos(x2)]" back in our formula, and up above in Step 2 I declared it to be = ln|y|

So I sub ln|y| into the formula and get...

ln|y| = cos(x2) + 2sin(x2) - 4*ln|y|
5*ln|y| = cos(x2) + 2sin(x2)
ln|y| = (1/5)*(cos(x2) + 2sin(x2))

y = e^(above line)

Ta da? lemmy know if this correct please!
 
Last edited:

Answers and Replies

  • #2
85
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integral of [ x.cos(x2) dx ] doesn't need integration by parts right? derivative of x square is 2 times x.
 
  • #3
25
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integral of [ x.cos(x2) dx ] doesn't need integration by parts right? derivative of x square is 2 times x.

Are you high? :P
What has the derivative of x^2 got to do with the integral of x*cos(2^x) ?
 
  • #4
302
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Why are you even using integration by parts? A much simpler approach is to substitute t=x^2 on your right hand integration.
 
  • #5
85
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HEY! I WASN'T HIGH! I was trying to help!
 
  • #6
25
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Why are you even using integration by parts? A much simpler approach is to substitute t=x^2 on your right hand integration.
As far as I'm aware the easiest way to integrate the product of two functions is to integrate it by parts.

Could you please explain your method? ie. Once I substitute t=x^2, what then?
 
  • #7
302
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If t=x^2, dt=2xdx. Your integral xcos(x2)dx can then be expressed as (1/2)cos(t)dt, which you can integrate easily.
 
  • #8
25
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If t=x^2, dt=2xdx. Your integral xcos(x2)dx can then be expressed as (1/2)cos(t)dt, which you can integrate easily.

How does x become (1/2) ?
 
  • #9
302
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You haven't done integrations by substitution yet? Maybe you should wait for your class to get to that point...(Actually if you are doing differential equations, you should be aware of this)

Anyway: xcos(x2)dx = (1/2)cos(x2)(2xdx) = (1/2)cos(t)dt
 
  • #10
SammyS
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Science Advisor
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How does x become (1/2) ?
This is pretty standard when using substitution to do integration. Generally the method of
substitution is covered before integration by parts.

You are trying to find [itex]\displaystyle \int x\cos(x^2)\,dx\,.[/itex]

As Oster suggested, use substitution. Pi-Bond gave more details: use t = x2, then dt = 2x dx → x dx = (1/2) dt . This is where the 1/2 comes from.

x doesn't become 1/2. Pi-Bond moved the x to be with 2 x dx, then had to include 1/2 to make up for the extra factor of 2. It's just a different way to get the same result.
 
  • #11
25
0
This is pretty standard when using substitution to do integration. Generally the method of
substitution is covered before integration by parts.

You are trying to find [itex]\displaystyle \int x\cos(x^2)\,dx\,.[/itex]

As Oster suggested, use substitution. Pi-Bond gave more details: use t = x2, then dt = 2x dx → x dx = (1/2) dt . This is where the 1/2 comes from.

x doesn't become 1/2. Pi-Bond moved the x to be with 2 x dx, then had to include 1/2 to make up for the extra factor of 2. It's just a different way to get the same result.

Thanks guys, understand it all now. Legends.
 

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