1. ## simple Integral problem

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

I know this is related to Inverse Tringemetric functions, so the Integral should somehow end $\displaystyle {sin^{-1}}x$

2. try $\displaystyle x = 3 \sin u$
Bobak

3. Sorry, are you meaning using $\displaystyle x = 3 \sin u$ for u substitution?
$\displaystyle \int \frac{{\cos u}}{\sqrt{9-3 \sin u}}$
I can't see how that works :s

4. Originally Posted by hlolli
Sorry, are you meaning using $\displaystyle x = 3 \sin u$ for u substitution?
$\displaystyle \int \frac{{\cos u}}{\sqrt{9-3 \sin u}}$
I can't see how that works :s
you made a little error. you didn't square x, also$\displaystyle dx = 3 \cos u du$

Bobak

5. $\displaystyle \int \frac{3{\cos u}du}{\sqrt{9-9 \sin^{2}u}}$

$\displaystyle \int \frac{1}{\sqrt{9-9 \sin^{2}u}}{3{\cos u}du}$

But please stay with me, this is in a chapter of Tringometric functions

and
$\displaystyle \frac{d}{dx} (\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$

This should somehow relate, now I'm stuck with 9 that I want to become 1

6. $\displaystyle 9-9 \sin^{2}u = 9(1-\sin^{2}u ) = ....$

Bobak

7. does sin^2 x + cos^2 x = 1 ring any bells?

8. It's solved Thanks!