MHB Integrate x/(4-x^4)^0.5: Solution

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The discussion focuses on integrating the function x/((4-x^4)^0.5). Several users attempt different substitution methods, including u = 4 - x^4 and u = x^2, but encounter difficulties. A key suggestion is to rewrite the integral in terms of u, specifically using u = x^2/2, to simplify the expression. Participants emphasize the importance of recognizing familiar integral forms to facilitate the solution. The conversation concludes with expressions of gratitude for the assistance provided.
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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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