By making the substitution u-(sin)^(-1) (x/3), simplify: x^2/√(9-x^2 )

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- Jul 24th 2011, 01:54 PMJamesBond16By making the substitution..
By making the substitution u-(sin)^(-1) (x/3), simplify: x^2/√(9-x^2 )

- Jul 24th 2011, 02:10 PMSironRe: By making the substitution..
Is the substitution:

$\displaystyle u=\sin^{-1}\cdot \frac{x}{3}$ or?

Because that would be strange. - Jul 24th 2011, 02:21 PMpickslidesRe: By making the substitution..
$\displaystyle \displaystyle u=\sin^{-1} \frac{x}{3}$

$\displaystyle \displaystyle \sin u= \frac{x}{3}$

$\displaystyle \displaystyle x = 3\sin u$

So the expression becomes

$\displaystyle \displaystyle \frac{x^2}{\sqrt{9-x^2}} = \frac{9\sin^2 u}{\sqrt{9-(3\sin u)^2}} = \dots$

Can you finish it from here?

Do you need to integrate this function? If so, consider $\displaystyle \displaystyle x = 3\sin u \implies dx = 3\cos u~du$ - Jul 24th 2011, 02:24 PMJamesBond16Re: By making the substitution..
yep thats it

- Jul 24th 2011, 02:25 PMJamesBond16Re: By making the substitution..
- Jul 24th 2011, 02:29 PMpickslidesRe: By making the substitution..
- Jul 24th 2011, 02:46 PMJamesBond16Re: By making the substitution..
[3 sin(u)]^2 / sqrt(9 - [3 sin(u)]^2)

[ 9 sin^2(u) ] / sqrt(9 - 9 sin^2(u))

[ 9 sin^2(u) ] / sqrt(9{1 - sin^2(u)})

[ 9 sin^2(u) ] / sqrt(9cos^2(u))

[ 9 sin^2(u) ] / [3 cos(u)]

[ 3 sin^2(u) ] / [ cos(u)]

3 tan(u) sin(u)

YEa? - Jul 24th 2011, 02:55 PMpickslidesRe: By making the substitution..
Nice work old mate!