# Thread: what is this anti derivative ?

1. ## what is this anti derivative ?

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

2. ## Re: what is this anti derivative ?

integrate 1/sqrt(5 - x^2) - Wolfram|Alpha ... click on 'show steps'.

Just in case a picture helps...

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

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

The general drift is...

__________________________________________________ ____________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

3. ## Re: 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????

4. ## Re: 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.

5. ## Re: what is this anti derivative ?

for constants $\displaystyle a$ , $\displaystyle b$ , and $\displaystyle c$ ...

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

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

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

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

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

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

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