1/sq root of (5-x^2) and how do you get there??

Printable View

- Jul 15th 2012, 09:10 AMsluggerbrothwhat is this anti derivative ?
1/sq root of (5-x^2) and how do you get there??

- Jul 15th 2012, 10:09 AMtom@ballooncalculusRe: what is this anti derivative ?
integrate 1/sqrt(5 - x^2) - Wolfram|Alpha ... click on 'show steps'.

Just in case a picture helps...

http://www.ballooncalculus.org/draw/...l/thirteen.png

... where (key in spoiler) ...

__Spoiler__:

__________________________________________________ ____________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jul 15th 2012, 10:34 AMsluggerbrothRe: what is this anti derivative ?
Mostly I wanted to see how you move from sq root of (5-x^2) to (1-x^2) in denominator. please show u=x/2 substitution and the algebra to get there????

- Jul 15th 2012, 10:50 AMtom@ballooncalculusRe: what is this anti derivative ?
Well I showed you the very first step. The next is to bring the 5 outside of the square root...

$\displaystyle \frac{1}{\sqrt{5}\sqrt{(1 - \frac{1}{5}x^2)}}$

Then bring the fifth inside of the square, with x, and bring the 5 in the denominator into its own fraction...

$\displaystyle \frac{1}{\sqrt{5}}\ \frac{1}{\sqrt{1 - (\frac{x}{\sqrt{5}})^2}}$

Now, you can substitute sin theta for x over root 5, which is what I've done.

Or you could sub root 5 sin theta for x. - Jul 15th 2012, 01:51 PMskeeterRe: what is this anti derivative ?
for constants $\displaystyle a$ , $\displaystyle b$ , and $\displaystyle c$ ...

$\displaystyle \int \frac{a}{\sqrt{b - cx^2}} \, dx$

$\displaystyle \int \frac{a}{\sqrt{b} \cdot \sqrt{1 - \frac{c}{b}x^2}} \, dx$

$\displaystyle \frac{a}{\sqrt{b}} \int \frac{1}{\sqrt{1 - \frac{c}{b}x^2}} \, dx$

$\displaystyle \frac{a}{\sqrt{b}} \int \frac{1}{\sqrt{1 - \left(\sqrt{\frac{c}{b}}x \right)^2}} \, dx$

$\displaystyle \sqrt{\frac{b}{c}} \cdot \frac{a}{\sqrt{b}} \int \frac{\sqrt{\frac{c}{b}}}{\sqrt{1 - \left(\sqrt{\frac{c}{b}}x \right)^2}} \, dx$

$\displaystyle \frac{a}{\sqrt{c}} \int \frac{\sqrt{\frac{c}{b}}}{\sqrt{1 - \left(\sqrt{\frac{c}{b}}x \right)^2}} \, dx$

$\displaystyle \frac{a}{\sqrt{c}} \arcsin\left(\sqrt{\frac{c}{b}}x \right) + C$