# Integral of 5x^2 /sqrt(4x-x^2) dx

s3a

## Homework Statement

Integral of 5x^2 /sqrt(4x-x^2) dx. (This can also be seen in the attached TheProblem.png file.)

## Homework Equations

Integration with variable substitution.

## The Attempt at a Solution

Could someone please tell me what I did wrong in my work (which is attached as MyWork.jpg)?

#### Attachments

• TheProblem.png
737 bytes · Views: 1,146
• MyWork.jpg
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Mentor

## Homework Statement

Integral of 5x^2 /sqrt(4x-x^2) dx. (This can also be seen in the attached TheProblem.png file.)

## Homework Equations

Integration with variable substitution.

## The Attempt at a Solution

Could someone please tell me what I did wrong in my work (which is attached as MyWork.jpg)?

You can't do what you did when you squared the integrand. IOW, in the step where the integrand changes from ## \frac{5x^{3/2}}{\sqrt{4 -x}}## to ##\frac{25x^3}{4 - x} ##.

iRaid
Ouch why would anyone ever make you solve this. You have to complete the square, then do a trig sub, followed by a bunch of manipulation of variables.

1 person
s3a
You can't do what you did when you squared the integrand. IOW, in the step where the integrand changes from ## \frac{5x^{3/2}}{\sqrt{4 -x}}## to ##\frac{25x^3}{4 - x} ##.
I don't know if this is answerable in a way other than “you just can't” but, why can't I do that?

Ouch why would anyone ever make you solve this. You have to complete the square, then do a trig sub, followed by a bunch of manipulation of variables.
[Dark Knight movie quote]Some men just want to see the world burn.[/Dark Knight movie quote] :P

But, on a serious note, are you saying to make what's under the root be -(x-2)^2 + 4?

Mentor
I don't know if this is answerable in a way other than “you just can't” but, why can't I do that?
Because, in general ##\int f(x)dx \neq \int [f(x)]^2 dx##. When you square the integrand, you get a different integrand.
But, on a serious note, are you saying to make what's under the root be -(x-2)^2 + 4?
That seems to be what iRaid was suggesting.

s3a
Because, in general ##\int f(x)dx \neq \int [f(x)]^2 dx##. When you square the integrand, you get a different integrand.
Oh, is it because squaring (or something similar) yields volumes?

That seems to be what iRaid was suggesting.
Alright. :) So, is my latest attachment correct so far?

#### Attachments

• MyWork.jpg
14.6 KB · Views: 1,627
Mentor
Oh, is it because squaring (or something similar) yields volumes?
No, nothing to do with that. If you square something, what you get is usually different. You wouldn't expect the integral to remain the same, would you?

Looking at it another way, do you think that d/dx(x2) = d/dx(x4)?
Alright. :) So, is my latest attachment correct so far?
You can check for yourself. If you differentiate your answer, you should get back your original integrand.

s3a
Looking at it another way, do you think that d/dx(x2) = d/dx(x4)?
No but, I think I see a better way of understanding it; 3/4 ≠ 3^2 / 4^2 = 9/16 so, integrating each yields a different answer (and, it's different from squaring both sides of an equation).

You can check for yourself. If you differentiate your answer, you should get back your original integrand.
I don't see how I can bring back the square root plus, Wolfram Alpha seems to imply that the two expressions are NOT equal … where have I gone wrong this time? It's probably something small like the squaring of the numerator and denominator from last time but, I just can't see what it is!

Mentor
In your latest work, it looks like you just swapped dx for dθ to get the new integral. If that's what you did, that's where you went wrong. If 2sinθ = x - 2, then dx = 2cosθdθ.

Also, in you latest work, you have 5x3 in the numerator - it should be 5x2.

After completing the square, you should have this integral:
$$\int \frac{5x^2 dx}{\sqrt{4 - (x - 2)^2}}$$

At this point I would bring the 5 outside the integral, and would do an ordinary substitution with u = x - 2. Then I would do the trig substitution, with 2sin(θ) = u and 2cos(θ)dθ = du.

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1 person
s3a
I now see how to do that problem! Thank you (both) very much!