How can I integrate x/(sqrrt(1-9x^2)?

Thanks

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- May 11th 2008, 07:33 PMdangerous_daveintegrate: x/(sqrrt(1-9x^2)
How can I integrate x/(sqrrt(1-9x^2)?

Thanks - May 11th 2008, 07:37 PMMathstud28
- May 11th 2008, 07:37 PMo_O
$\displaystyle {\color{blue}u = 1 -9x^{2}}$

$\displaystyle du = -18x dx \Rightarrow {\color{red}x dx = -\frac{du}{18}}$

Looking at your integral:

$\displaystyle \int \frac{{\color{red}xdx}}{\sqrt{{\color{blue}1-9x^{2}}}}$ - May 11th 2008, 07:38 PMMathstud28
- May 11th 2008, 07:38 PMo_O
Haha uh ... let's call it a tie :p

- May 11th 2008, 07:40 PMMathstud28
- May 11th 2008, 07:42 PMChris L T521
To integrate $\displaystyle \int \frac{x}{\sqrt{1-9x^2}}\,dx$, you would need to make a u-substitution: $\displaystyle u=1-9x^2$. As a result, $\displaystyle du=-18x\,dx$. Let's substitute all this stuff into the integral now:

$\displaystyle \int \frac{x}{\sqrt{1-9x^2}}\,dx$ now becomes $\displaystyle -\frac{1}{18}\int\frac{\,du}{\sqrt{u}}=-\frac{1}{18}\int u^{-\frac{1}{2}}\,du$.

Integrate:

$\displaystyle -\frac{1}{18}\int u^{-\frac{1}{2}}\,du=-\frac{1}{9}\sqrt{u}+C$.

Substitute u back in and we have:

$\displaystyle \int \frac{x}{\sqrt{1-9x^2}}\,dx=-\frac{1}{9}\sqrt{1-9x^2}+C$

Hope that helped you out!!! :D - May 11th 2008, 07:57 PMdangerous_dave
That does help heaps, thanks :D