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

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

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- Dec 15th 2008, 06:54 AMhlollisimple 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 $\displaystyle {sin^{-1}}x$ - Dec 15th 2008, 06:56 AMbobak
try $\displaystyle x = 3 \sin u $

Bobak - Dec 15th 2008, 07:03 AMhlolli
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 - Dec 15th 2008, 07:06 AMbobak
- Dec 15th 2008, 07:34 AMhlolli
$\displaystyle \int \frac{3{\cos u}du}{\sqrt{9-9 \sin^{2}u}}$

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

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 - Dec 15th 2008, 07:53 AMbobak
$\displaystyle 9-9 \sin^{2}u = 9(1-\sin^{2}u ) = ....$

Bobak - Dec 15th 2008, 07:56 AMlordcrusade9
does sin^2 x + cos^2 x = 1 ring any bells?

- Dec 15th 2008, 08:07 AMhlolli
It's solved Thanks!