# Find the integral

#### renyikouniao

##### Member
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 theres x at the top.

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Use the subtitution $$\displaystyle u = x^2$$

#### renyikouniao

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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

##### Member
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?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I tried to solve this using Integral1/((1-x^2)^0.5)=sin^-1(x)
You can use this , right ? , but how ?

#### renyikouniao

##### Member
You can use this , right ? , but how ?
No,we can't.That's the problem

#### renyikouniao

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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

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

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)

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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

##### Member
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 $$\displaystyle u = \frac{x^2}{2}$$
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