Integrate x/(4-x^4)^0.5: Solution

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The integral of the function x/((4-x^4)^0.5) can be simplified using the substitution u = x^2/2. This transformation leads to the expression 1/(2(4-u^2)^0.5) du, allowing for further integration techniques. The discussion emphasizes the importance of recognizing patterns in integrals and utilizing appropriate substitutions to facilitate the integration process. Participants also noted that alternative methods, such as using arcsine, were not permitted in their context.

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Question:Integrate x/((4-x^4)^0.5)

I tried to solve this using Integral1/((1-x^2)^0.5)=sin^-1(x)
But it didn't work out since there's x at the top.

And then I tried using u=4-x^4 ,It didn't work out neither
 
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Use the subtitution $$u = x^2$$
 
ZaidAlyafey said:
Use the subtitution $$u = x^2$$

Then the integral becomes 1/(2(4-u^2)^0.5) du?How to simplify this?
 
renyikouniao said:
Then the integral becomes 1/(2(4-u^2)^0.5) du?How to simplify this?

Doesn't that look familiar ? , just a little modification .
 
ZaidAlyafey said:
Doesn't that look familiar ? , just a little modification .

If you mean integral (1/((a^2-x^2)^0.5)=sin^-1(x/a)+c?Our professor doesn't want us ues this.Do you have any other method?
 
renyikouniao said:
I tried to solve this using Integral1/((1-x^2)^0.5)=sin^-1(x)

You can use this , right ? , but how ?
 
ZaidAlyafey said:
You can use this , right ? , but how ?

No,we can't.That's the problem
 
ZaidAlyafey said:
You can use this , right ? , but how ?

Do you have any suggestions on how to solve this?:confused:
 
renyikouniao said:
Do you have any suggestions on how to solve this?:confused:

$$\frac{1}{\sqrt{4-x^2}} = \frac{1}{2\sqrt{1-\left(\frac{x}{2}\right)^2}}$$

Now you can use the substitution $$u = \frac{x}{2}$$
 
  • #10
ZaidAlyafey said:
$$\frac{1}{\sqrt{4-x^2}} = \frac{1}{2\sqrt{1-\left(\frac{x}{2}\right)^2}}$$

Now you can use the substitution $$u = \frac{x}{2}$$

Thank you;),but what about the x on the top,should I rewrite x=2u?But if I do so,I can't use integral 1/((1-x^2)^0.5) right?

Integrate x/((4-x^4)^0.5)
 
  • #11
renyikouniao said:
Then the integral becomes 1/(2(4-u^2)^0.5) du?How to simplify this?

You already arrive to this part . you can do the little trick I provided .

otherwise

$$\frac{x}{\sqrt{4-x^4}} = \frac{x}{2 \sqrt{1-\left( \frac{x^2}{2} \right)^2}}$$

you can know make the subtitution $$u = \frac{x^2}{2}$$
 
  • #12
ZaidAlyafey said:
You already arrive to this part . you can do the little trick I provided .

otherwise

$$\frac{x}{\sqrt{4-x^4}} = \frac{x}{2 \sqrt{1-\left( \frac{x^2}{2} \right)^2}}$$

you can know make the subtitution $$u = \frac{x^2}{2}$$

Thank you very much for you patients(flower)(flower)(flower)
 

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