# Thread: Simple Fraction Integration

1. ## Simple Fraction Integration

Q: $\displaystyle \displaystyle\int \frac{x}{\sqrt{4-x^2}} \, \mathrm{d}x$

Does that just equal $\displaystyle \sqrt{4-x^2}$? If so, why?

2. Originally Posted by Air
Q: $\displaystyle \displaystyle\int \frac{x}{\sqrt{4-x^2}} \, \mathrm{d}x$

Does that just equal $\displaystyle \sqrt{4-x^2}$? If so, why?

Let $\displaystyle u = 4 - x^2$

Then $\displaystyle \frac{du}{dx} = -2x$

$\displaystyle \int \frac{x}{\sqrt{4-x^2}} dx$

$\displaystyle = -\frac{1}{2} \int \frac{1}{\sqrt{u}} du$

$\displaystyle = -\frac{1}{2} \int (u)^{- \frac{1}{2}} du$

$\displaystyle = -\frac{1}{2} (2(u)^{\frac{1}{2}})$

$\displaystyle = -(\sqrt{4 - x^2})$

3. Hello,

Originally Posted by Air
Q: $\displaystyle \displaystyle\int \frac{x}{\sqrt{4-x^2}} \, \mathrm{d}x$

Does that just equal $\displaystyle \sqrt{4-x^2}$? If so, why?
$\displaystyle \frac{x}{\sqrt{4-x^2}}=\frac{2x}{2 \sqrt{4-x^2}}$

Here, it's like $\displaystyle \frac{-u'(x)}{2 \sqrt{u(x)}}$

An antiderivative of this is $\displaystyle -\sqrt{u(x)}=-\sqrt{4-x^2}$

4. When I use substitution of $\displaystyle u^2 = 4 - x^2$, I seem to get an integral of $\displaystyle \int - \frac{1}{u} \ \mathrm{d}u$, but that would be $\displaystyle \ln u$. Doesn't $\displaystyle u^2 = 4 - x^2$ substitution work?

5. Originally Posted by Air
When I use substitution of $\displaystyle u^2 = 4 - x^2$, I seem to get an integral of $\displaystyle \int - \frac{1}{u} \ \mathrm{d}u$, but that would be $\displaystyle \ln u$. Doesn't $\displaystyle u^2 = 4 - x^2$ substitution work?
If you do this substitution :

$\displaystyle 2 udu=-2x dx$ -> $\displaystyle dx=-\frac ux$

-> $\displaystyle \int \frac{x}{\sqrt{4-x^2}} dx=\int \frac{u}{u} du=u=\pm \sqrt{4-x^2}$

It's quite dangerous because of this $\displaystyle \pm$

6. Originally Posted by Moo
If you do this substitution :

$\displaystyle 2 udu=-2x dx$ -> $\displaystyle dx=-\frac ux$

-> $\displaystyle \int \frac{x}{\sqrt{4-x^2}} dx=\int \frac{u}{u} du=u=\pm \sqrt{4-x^2}$

It's quite dangerous because of this $\displaystyle \pm$
Why did you get $\displaystyle \int \frac{u}{u} \ \mathrm{d}u$? Where did the $\displaystyle u$ in the numerator come from?

7. Originally Posted by Air
Why did you get $\displaystyle \int \frac{u}{u} \ \mathrm{d}u$? Where did the $\displaystyle u$ in the numerator come from?
From dx

$\displaystyle \int \frac{x}{\sqrt{4-x^2}} \underbrace{dx}_{\frac{-u}{x} \cdot du}=\int \frac{x}{u} \cdot \frac{-u}{x} du$

Typo
Originally Posted by Moo
If you do this substitution :

$\displaystyle 2 udu=-2x dx$ -> $\displaystyle dx=-\frac ux$

-> $\displaystyle \int \frac{x}{\sqrt{4-x^2}} dx={\color{red}-}\int \frac{u}{u} du={\color{red}-}u=\pm \sqrt{4-x^2}$

It's quite dangerous because of this $\displaystyle \pm$