1. ## integrate: x/(sqrrt(1-9x^2)

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

Thanks

2. Originally Posted by dangerous_dave
How can I integrate x/(sqrrt(1-9x^2)?

Thanks
You can do this two ways the first is this...recognize that

$\displaystyle \int\frac{x}{\sqrt{1-9x^2}}dx=\frac{-1}{9}\int\bigg(\sqrt{1-9x^2}\bigg)'dx$ where the answer becoems apparent

The other way is u-sub

Let $\displaystyle u=1-9x^2$

3. $\displaystyle {\color{blue}u = 1 -9x^{2}}$
$\displaystyle du = -18x dx \Rightarrow {\color{red}x dx = -\frac{du}{18}}$

$\displaystyle \int \frac{{\color{red}xdx}}{\sqrt{{\color{blue}1-9x^{2}}}}$

4. Originally Posted by o_O
$\displaystyle {\color{blue}u = 1 -9x^{2}}$
$\displaystyle du = -18x dx \Rightarrow {\color{red}x dx = -\frac{du}{18}}$

$\displaystyle \int \frac{{\color{red}xdx}}{\sqrt{{\color{blue}1-9x^{2}}}}$
Ahh dear o_O whats the score again?

5. Haha uh ... let's call it a tie

6. Originally Posted by o_O
Haha uh ... let's call it a tie
Sounds great to me...you are my team leader after all ...::quivers in fear::

7. Originally Posted by dangerous_dave
How can I integrate x/(sqrrt(1-9x^2)?

Thanks
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!!!

8. That does help heaps, thanks