Integration by parts involving square root!

  • #1

Homework Statement


|x3sqrt(4-x2)dx


Homework Equations


uv - | vdu


The Attempt at a Solution


u = x2 v = -1/3(4-x2)3/2
du = -2xdx dv = x(4-x2)1/2

uv - | vdu

x2(1/3)(4-x2)3/2 - | 1/3(4-x2)3/2(2xdx)
x2(1/3)(4-x2)3/2 +(1/3)|(4-x2)3/2(2xdx)

u = 4 - x2
du = -2xdx

x2(1/3)(4-x2)3/2 - (1/3)| u3/2du
x2(1/3)(4-x2)3/2-(1/3)2/5u5/2
x2(1/3)(4-x2)3/2-(1/3)(2/5)(4-x2)5/2
x2-(1/3)(4-x2)3/2-(2/15)(4-x2)5/2

Is this all I need? The answer I was supposed to get is -1/15(4-x2)3/2(8+3x2) I guess I dont see how that correlates
 
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Answers and Replies

  • #2
SammyS
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Hello learnonthefly. Welcome to PF !

Homework Statement


∫x^3sqrt(4-x^2)dx

Homework Equations


uv - ∫ vdu

The Attempt at a Solution


u = x^2 v = (-)1/3(4-x^2)^3/2
du = 2xdx dv = x(4-x^2)^1/2 dx

uv - ∫ vdu
I made a correction & did some editing above.

x^2(1/3)(4-x^2)^3/2 - | 1/3(4-x)^3/2(2xdx)
x^2(1/3)(4-x^2)^3/2 -(1/3)|(4-x^2)^3/2(2xdx)

u = 4 - x^2
du = 2xdx

x^2(1/3)(4-x^2)^3/2 - (1/3)| u^3/2du
x^2(1/3)(4-x^2)^3/2-(1/3)2/5u^5/2
x^2(1/3)(4-x^2)^3/2-(1/3)(2/5)(4-x^2)^5/2
That is kind of hard to read.

What is the question that you have for us ?
 
  • #3
Curious3141
Homework Helper
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87

Homework Statement


|x^3sqrt(4-x^2)dx


Homework Equations


uv - | vdu


The Attempt at a Solution


u = x^2 v = 1/3(4-x^2)^3/2
du = 2xdx dv = x(4-x^2)^1/2

uv - | vdu

x^2(1/3)(4-x^2)^3/2 - | 1/3(4-x)^3/2(2xdx)
x^2(1/3)(4-x^2)^3/2 -(1/3)|(4-x^2)^3/2(2xdx)

u = 4 - x^2
du = 2xdx

x^2(1/3)(4-x^2)^3/2 - (1/3)| u^3/2du
x^2(1/3)(4-x^2)^3/2-(1/3)2/5u^5/2
x^2(1/3)(4-x^2)^3/2-(1/3)(2/5)(4-x^2)^5/2

Homework Statement





Homework Equations





The Attempt at a Solution

You didn't actually ask a question. And it's tough to read what you've written - please format in Latex, it makes things a lot easier. :smile:

At any rate, your final answer is wrong, but it's probably a simple sign error.

For example, in the first integration by parts, ##v = -\frac{1}{3}{(4-x^2)}^{\frac{3}{2}}##. This is because of the ##-x^2## term within the parentheses.

You should go through your working carefully looking at all the sign errors and correct them.

EDIT: beaten by SammyS :biggrin:
 
  • #4
Sorry guys, updated and cleaned
 
  • #5
SammyS
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...

Is this all I need? The answer I was supposed to get is -1/15(4-x2)3/2(8+3x2) I guess I dont see how that correlates
Your result is equivalent to:

[itex]\displaystyle (1/15) (4-x^2)^{3/2} (3 x^2+8)[/itex]

You have a sign error in you integration by substitution part.
 
  • #6
Yea I see where I messed up I fixed that above. Could the algebraically inclined show me the manipulation to get the answer?
 
  • #7
SammyS
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Yea I see where I messed up I fixed that above. Could the algebraically inclined show me the manipulation to get the answer?
Factor [itex]\displaystyle \ (4-x^2)^{3/2}\ [/itex] from your result & collect terms.
 
  • #8
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I think an easier way to do this would be make u=x^2 right from the beginning. Then after that make another substitution y=4-u. Then it's pretty easy from there.
 

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