Help with trig sub integral

[Kristal9]

I've gone over my work twice, and while I dropped a negative the first time, I believe that my work is sound. Webwork is not accepting it, however.

Here is the problem and my solution. Sorry for bad handwriting

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[Romsek]

I have no idea from this what the original integral was.

 

[Kristal9]

I have no idea from this what the original integral was.

The original problem is in the top left. Sorry it took so long for this to show. It has to be approved first.

 

[Romsek]

[MATH]7 \displaystyle \int \dfrac{x^2}{4x - x^2}~dx[/MATH]
Is that it?

 

[Steven G]

It was a bit hard to follow as it looked like at the left side a negative sign appeared.
Why not take the derivative of your answer and see if you get back the integrand?
Also you are not finished. You should know what 56/2 equals and you should do the subtraction inside the parenthesis.

 

[pka]

I think that this is the basic integral, \(\displaystyle\int {\dfrac{{{x^2}}}{{\sqrt {4x - {x^2}} }}dx} \) please correct me if I am mistaken.
Otherwise, have a look at this.

 

[Kristal9]

I think that this is the basic integral, \(\displaystyle\int {\dfrac{{{x^2}}}{{\sqrt {4x - {x^2}} }}dx} \) please correct me if I am mistaken.
Otherwise, have a look at this.

Yes, that's right. Also, I did look at old reliable there, but I wanted to be able to do it for myself. There are many ways to solve an integral, and it isn't clear to me how it got that solution. It did some algebraic shuffling, it looks like, and anyway, the topic this week is trig integrals.

I also asked on math stackexchange here. Following Ans4's start I got this:

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I have one chance to finish this problem and am 100% on all webwork psets to this point, so a yes or no as to the correctness of the answer would be much appreciated. Also perhaps some guidance, if necessary.

 

[Kristal9]

Yea, obviously I was/am getting tired. My post above is still awaiting moderator clearance though it was made 90 minutes ago, and I also cannot edit it since too much time has elapsed.

Here are the edits I would have put in:

Here is the link to the other thread on math stackexchange (which I neglected to even put in):

Help with trig sub integral

I've gone over my work twice, and while I dropped a negative the first time, I believe that my work is sound. Webwork is not accepting it, however. Here is the problem and my solution. Sorry for bad

math.stackexchange.com

And here is the reworked answer I have:

< Link to "Most Viral Images" at imgur removed >

Obviously I could just put in a result from one of the more than one online integral calculators, but I am trying to do this myself.

If anyone would like to let me know if I have done fine, it would be nice.

Also, if anyone would like to let me know why the sub in my original work doesn't work, then that would also be nice.

 

[Deleted member 4993]

\(\displaystyle \int \frac{x^2}{\sqrt{4x - x^2}} dx \ \)

= \(\displaystyle \int \frac{x^2}{\sqrt{-4 + 4x - x^2 + 4}} dx \ \) ..................completing square [edited -4x to +4x]

= \(\displaystyle \int \frac{x^2}{\sqrt{2^2 - (x-2)^2 }} dx \ \) [edited]

substitute:

u = x-2

= \(\displaystyle \int \frac{(u+2)^2}{\sqrt{2^2 - u^2 }} du \ \)

substitute:

u = 2 * sin(v) ............ and ........... du = 2 * cos(v) dv

continue......

 

[Cubist]

Also, if anyone would like to let me know why the sub in my original work doesn't work, then that would also be nice.

I can see one reason why your original post didn't work. You proposed this sub:-

[math] 2\sin\left(\theta\right)=\sqrt{x} [/math]
And after differentiating you wrote:-

[math] 2\cos\left(\theta\right)=\mathrm{d}x [/math]
That is not correct - because you didn't take the sqrt into account

[math] \frac{\mathrm{d}}{\mathrm{d}\theta}2\sin\left(\theta\right)=\frac{\mathrm{d}}{\mathrm{d}\theta}\sqrt{x} [/math]
[math] 2\cos\left(\theta\right)= \frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{x}\right) \cdot \frac{\mathrm{d}x}{\mathrm{d}\theta} [/math]
[math] 2\cos\left(\theta\right)=\frac{1}{2\sqrt{x}} \cdot \frac{\mathrm{d}x}{\mathrm{d}\theta} [/math]
[math] 4\cos\left(\theta\right) \mathrm{d}\theta= \frac{1}{\sqrt{x}} \mathrm{d}x [/math]

 

[Cubist]

= \(\displaystyle \int \frac{x^2}{\sqrt{-4 - 4x - x^2 + 4}} dx \ \) ..................completing square

A slight sign mistake on the "-4x", I think it ought to be +"4x". Therefore:

= \(\displaystyle \int \frac{x^2}{\sqrt{2^2 - (x - 2)^2 }} dx \ \)

substitute: u = x - 2

= \(\displaystyle \int \frac{(u+2)^2}{\sqrt{2^2 - u^2 }} du \ \)

substitute:

u = 2 * sin(v) ............ and ........... du = 2 * cos(v) dv

continue......

 

[Cubist]

I continued Subhotosh Khan's method on from post#11 and got

Spoiler

[math] 6v-8\cos\left(v\right)-\sin\left(2v\right)+c [/math]
[math]=6\sin^{-1}\left(\frac{u}{2}\right)-\sqrt{1-\frac{u^2}{4}}\cdot \left(8+u\right)+c [/math]
[math]=6\sin^{-1}\left(\frac{x-2}{2}\right)-\left(6+x\right) \sqrt{x-\frac{x^2}{4}}+c[/math]
(This ignores the 7* outside the original integral expression)

I'm pretty sure the above is correct (plotting its numeric derivative agrees with the original)

OP's post#8 seems to be on the right track but not quite there

The Wolfam result, post#6, was incorrect - it feels good to beat the machine!

 

[Kristal9]

I can see one reason why your original post didn't work. You proposed this sub:-

[math] 2\sin\left(\theta\right)=\sqrt{x} [/math]
And after differentiating you wrote:-

[math] 2\cos\left(\theta\right)=\mathrm{d}x [/math]
That is not correct - because you didn't take the sqrt into account

[math] \frac{\mathrm{d}}{\mathrm{d}\theta}2\sin\left(\theta\right)=\frac{\mathrm{d}}{\mathrm{d}\theta}\sqrt{x} [/math]
[math] 2\cos\left(\theta\right)= \frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{x}\right) \cdot \frac{\mathrm{d}x}{\mathrm{d}\theta} [/math]
[math] 2\cos\left(\theta\right)=\frac{1}{2\sqrt{x}} \cdot \frac{\mathrm{d}x}{\mathrm{d}\theta} [/math]
[math] 4\cos\left(\theta\right) \mathrm{d}\theta= \frac{1}{\sqrt{x}} \mathrm{d}x [/math]

Thanks

 

[Deleted member 4993]

A slight sign mistake on the "-4x", I think it ought to be +"4x". Therefore:

= \(\displaystyle \int \frac{x^2}{\sqrt{2^2 - (x - 2)^2 }} dx \ \)

substitute: u = x - 2

= \(\displaystyle \int \frac{(u+2)^2}{\sqrt{2^2 - u^2 }} du \ \)

Note from "banished to the corner":

You are correct and I edited my post.