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#### renyikouniao

##### Member

- Jun 1, 2013

- 41

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

But it didn't work out since theres x at the top.

And then I tried using u=4-x^4 ,It didn't work out neither

- Thread starter renyikouniao
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- Thread starter
- #1

- Jun 1, 2013

- 41

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

But it didn't work out since theres x at the top.

And then I tried using u=4-x^4 ,It didn't work out neither

- Jan 17, 2013

- 1,667

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

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- #3

- Jun 1, 2013

- 41

Then the integral becomes 1/(2(4-u^2)^0.5) du?How to simplify this?Use the subtitution \(\displaystyle u = x^2\)

- Jan 17, 2013

- 1,667

Doesn't that look familiar ? , just a little modification .Then the integral becomes 1/(2(4-u^2)^0.5) du?How to simplify this?

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- #5

- Jun 1, 2013

- 41

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

- Jan 17, 2013

- 1,667

You can use this , right ? , but how ?I tried to solve this using Integral1/((1-x^2)^0.5)=sin^-1(x)

- Thread starter
- #7

- Jun 1, 2013

- 41

No,we can't.That's the problemYou can use this , right ? , but how ?

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- #8

- Jun 1, 2013

- 41

Do you have any suggestions on how to solve this?You can use this , right ? , but how ?

- Jan 17, 2013

- 1,667

\(\displaystyle \frac{1}{\sqrt{4-x^2}} = \frac{1}{2\sqrt{1-\left(\frac{x}{2}\right)^2}}\)Do you have any suggestions on how to solve this?

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

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- #10

- Jun 1, 2013

- 41

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

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

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

- Jan 17, 2013

- 1,667

You already arrive to this part . you can do the little trick I provided .Then the integral becomes 1/(2(4-u^2)^0.5) du?How to simplify this?

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 \(\displaystyle u = \frac{x^2}{2}\)

- Thread starter
- #12

- Jun 1, 2013

- 41

Thank you very much for you patients(flower)(flower)(flower)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 \(\displaystyle u = \frac{x^2}{2}\)