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

[renyikouniao]

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

 

[alyafey22]

Use the subtitution \(\displaystyle u = x^2\)

 

[renyikouniao]

Use the subtitution \(\displaystyle u = x^2\)

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

 

[alyafey22]

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 .

 

[renyikouniao]

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?

 

[alyafey22]

I tried to solve this using Integral1/((1-x^2)^0.5)=sin^-1(x)

You can use this , right ? , but how ?

 

[renyikouniao]

You can use this , right ? , but how ?

No,we can't.That's the problem

 

[renyikouniao]

You can use this , right ? , but how ?

Do you have any suggestions on how to solve this?

 

[alyafey22]

Do you have any suggestions on how to solve this?

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

Now you can use the substitution \(\displaystyle u = \frac{x}{2}\)

 

[renyikouniao]

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

Now you can use the substitution \(\displaystyle u = \frac{x}{2}\)

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)

 

[alyafey22]

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

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

you can know make the subtitution \(\displaystyle u = \frac{x^2}{2}\)

 

[renyikouniao]

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

otherwise

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

you can know make the subtitution \(\displaystyle u = \frac{x^2}{2}\)

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