simple Integral problem

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December 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 ${sin^{-1}}x$
December 15th 2008, 06:56 AM
bobak

try $x = 3 \sin u$
Bobak
December 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
December 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
December 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
December 15th 2008, 07:53 AM
bobak

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

Bobak
December 15th 2008, 07:56 AM
lordcrusade9

does sin^2 x + cos^2 x = 1 ring any bells?
December 15th 2008, 08:07 AM
hlolli

It's solved Thanks!

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