# simple Integral problem

• Dec 15th 2008, 06:54 AM
hlolli
simple Integral problem
$\int \frac{dx}{\sqrt{9-x^2}}$

I know this is related to Inverse Tringemetric functions, so the Integral should somehow end ${sin^{-1}}x$
• Dec 15th 2008, 06:56 AM
bobak
try $x = 3 \sin u$
Bobak
• Dec 15th 2008, 07:03 AM
hlolli
Sorry, are you meaning using $x = 3 \sin u$ for u substitution?
$\int \frac{{\cos u}}{\sqrt{9-3 \sin u}}$
I can't see how that works :s
• Dec 15th 2008, 07:06 AM
bobak
Quote:

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

you made a little error. you didn't square x, also $dx = 3 \cos u du$

Bobak
• Dec 15th 2008, 07:34 AM
hlolli
$\int \frac{3{\cos u}du}{\sqrt{9-9 \sin^{2}u}}$

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

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

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

This should somehow relate, now I'm stuck with 9 that I want to become 1
• Dec 15th 2008, 07:53 AM
bobak
$9-9 \sin^{2}u = 9(1-\sin^{2}u ) = ....$

Bobak
• Dec 15th 2008, 07:56 AM