Integration of x^2 dx / (1-x)^1/2

In summary, U substitution in an equation for x^2dx results in-2(1-x)^2+1/u. The attempted solution was to first find u and then use that to integrate the equation for x^2dx. However, when differentiation was attempted, x^2/sqrt(1-x) was not found.
  • #1
ganondorf29
54
0

Homework Statement



Integrate: x^2 dx / (1-x)^1/2

Homework Equations



U substitution

The Attempt at a Solution



First I defined u = (1-x)^1/2
du = -dx/2(1-x)^1/2
dx = -2(1-x)^1/2 du

then for x
u^2 = 1-x
x = 1-u^2

integral of: (1-u^2)^2/u
= (u^4-2u^2+1) / u
= u^3-2u+1/u

integrating:

u^4/4 - u^2 + 1/u

= ((1-x)^2)/4 - (1-x) + ln|1-x| + CIs this right?
 
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  • #2
You can try differentiating your answer to check if it is correct..
 
  • #3
I differentiated and the answer I got was wrong. Does anyone know what I did wrong?
 
  • #4
looks like you should have gotten ,

ln| (1-x)^1/2 |

EDIT: I was wrong here, look at Dick's hint later. It seems pretty easy after his hint.
 
Last edited:
  • #5
You figured out that dx=-2u*du. But then you never used it. There's no logs in the problem at all.
 
  • #6
If dx = -2u*du that means that I would have to set u = (1-x)^1/2 and square both sides and then take the derivative, right? Well, if I do that, what do I do for the x^2 on the numerator?
 
  • #7
ganondorf29 said:
If dx = -2u*du that means that I would have to set u = (1-x)^1/2 and square both sides and then take the derivative, right? Well, if I do that, what do I do for the x^2 on the numerator?

What?? No, you are doing everything right, but you forgot to replace the dx with an expression for du. You said "dx = -2(1-x)^1/2 du". Use that expression.
 
  • #8
Ok, I reworked the dx part and I got

Integral of:
((1-u^2)^2*-2u*du)/u
= -2(1-u^2)^2
= -2(u^4-2u^2+1)

= -(2/5)(u^5) + (4/3)(u^3) - 2u + C

= -(2/5)((1-x)^5) + (4/3)((1-x)^3) - 2(1-x) + C

But when I differentiate this I am not getting x^2/sqrt(1-x)
:confused:
 
  • #9
u=sqrt(1-x). You substituted u=(1-x).
 
  • #10
After fixing the substitution

-(2/5)((sqrt(1-x))^5) + (4/3)((sqrt(1-x))^3) - 2(sqrt(1-x)) + C

d/dx of -(2/5)((sqrt(1-x))^5) + (4/3)((sqrt(1-x))^3) - 2(sqrt(1-x)) + C

= 1/sqrt(1-x) - (x+1)*(sqrt(1-x))

which isn't x^2/(sqrt(1-x))


I went to mathematica's online integral calculator and they gave this:
-(2/15)*(sqrt(1-x)*(x(3x+4)+8

My Ti-89 gave the same answer once factored. I still don't know what I'm doing wrong.
 
  • #11
1/sqrt(1-x) - (x+1)*(sqrt(1-x)). Multiply the numerator and denominator of the second term by sqrt(1-x). 1/sqrt(1-x)-(x+1)(1-x)/sqrt(1-x)=(1-(1-x^2))/sqrt(1-x)=x^2/sqrt(1-x). It's right. You've got too many machines telling you what to do.
 
  • #12
Thank you so much
 

Related to Integration of x^2 dx / (1-x)^1/2

1. What is the formula for integrating x^2 dx / (1-x)^1/2?

The formula for integrating x^2 dx / (1-x)^1/2 is ∫(x^2 / (1-x)^1/2)dx.

2. How do I solve this integration using substitution?

To solve this integration using substitution, you can let u = 1-x. Then, dx = -du and the integral becomes ∫((-u+1)^2 / u^1/2)(-du). Expand the numerator and simplify the integral before solving.

3. Can this integration be solved using integration by parts?

Yes, this integration can be solved using integration by parts. Let u = (1-x)^1/2 and dv = x^2 dx. Then, du = -1/2(1-x)^-1/2 dx and v = 1/3x^3. The integral becomes ∫u dv = uv - ∫v du.

4. Is there an easier way to solve this integration?

Yes, there is. You can use the trigonometric substitution x = sin^2θ to solve this integration. The integral becomes ∫(sin^4θ cos^-3θ)dx. Use trigonometric identities to simplify the integral before solving.

5. What is the significance of integrating x^2 dx / (1-x)^1/2?

The integration of x^2 dx / (1-x)^1/2 is commonly used in physics and engineering to calculate the work done by a varying force over a certain distance. It is also used in statistics to find the probability density function of certain distributions, such as the Chi-square distribution.

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