# Integration: 1/sqrt(9-4x^2)

• Jun 1st 2009, 10:48 AM
calc101
Integration: 1/sqrt(9-4x^2)
I am doing some integration drills and I am stuck with this one:

$
\int \frac {1}{\sqrt{9-4x^2}} dx
$

the solution is as follows:

$u = \frac {2}{3}x$

$du = \frac {2}{3} dx$

Step 3: $\frac {1}{3} \int \frac {1}{\sqrt{1 - (\frac {2x}{3})^2}} dx$

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

$
= \frac {1}{2} sin^-1 (\frac {2}{3}x) + C
$

What I don't understand is why $\frac {2x}{3}$ was chosen as u? and how they got step number 3?

Any help is appreciated.
• Jun 1st 2009, 11:07 AM
Random Variable
Your'e missing something:

$\frac {1}{\sqrt{9-4x^{2}}} = \frac {1}{\sqrt{9 - (2x)^{2}}} = \frac {1}{3 \sqrt{1-(\frac {2x}{3})^{2}}}$
• Jun 1st 2009, 11:10 AM
Spec
$\int \frac{1}{\sqrt{9-4x^2}}dx=\int \frac{1}{3\sqrt{1-\frac{4x^2}{9}}}dx=\int \frac{1}{3\sqrt{1-\left(\frac{2x}{3}\right)^2}}dx$

$\frac{d}{du}\arcsin u=\frac{1}{\sqrt{1-u^2}}$
• Jun 1st 2009, 11:31 AM
calc101
I am sorry, I missed the $\frac {1}{3}$

Random Variable, I dont see how you got to the last step, could you elaborate?

So, I understand this:

$
\int \frac {1}{\sqrt{(3)^2 - (2x)^2}}
$

but to jump from that to this:

$
\frac {1}{3 \sqrt{1-(\frac {2x}{3})^{2}}}
$

is what I do not understand?
• Jun 1st 2009, 11:46 AM
Random Variable
Quote:

Originally Posted by calc101
I am sorry, I missed the $\frac {1}{3}$

Random Variable, I dont see how you got to the last step, could you elaborate?

So, I understand this:

$
\int \frac {1}{\sqrt{(3)^2 - (2x)^2}}
$

but to jump from that to this:

$
\frac {1}{3 \sqrt{1-(\frac {2x}{3})^{2}}}
$

is what I do not understand?

$\frac {1}{\sqrt{9 - (2x)^{2}}} = \frac {1}{\sqrt{9(1 - \frac {(2x)^{2}}{9}})} = \frac {1}{3 \sqrt{1 - (\frac {2x}{3})^{2}}}$