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

**dangerous_dave** · May 11th 2008, 07:33 PM

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

Thanks

**Mathstud28** · May 11th 2008, 07:37 PM

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

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

The other way is u-sub

Let $u=1-9x^2$

**o_O** · May 11th 2008, 07:37 PM

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

Looking at your integral:
$\int \frac{{\color{red}xdx}}{\sqrt{{\color{blue}1-9x^{2}}}}$

**Mathstud28** · May 11th 2008, 07:38 PM

Originally Posted by o_O

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

Looking at your integral:
$\int \frac{{\color{red}xdx}}{\sqrt{{\color{blue}1-9x^{2}}}}$

Ahh dear o_O whats the score again?

**o_O** · May 11th 2008, 07:38 PM

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

**Mathstud28** · May 11th 2008, 07:40 PM

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::

**Chris L T521** · May 11th 2008, 07:42 PM

Originally Posted by dangerous_dave

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

Thanks

To integrate $\int \frac{x}{\sqrt{1-9x^2}}\,dx$ , you would need to make a u-substitution: $u=1-9x^2$ . As a result, $du=-18x\,dx$ . Let's substitute all this stuff into the integral now:

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

Integrate:

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

Substitute u back in and we have:

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

Hope that helped you out!!!

**dangerous_dave** · May 11th 2008, 07:57 PM

That does help heaps, thanks

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