By making the substitution..

**JamesBond16** · July 24th 2011, 01:54 PM

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

**Siron** · July 24th 2011, 02:10 PM

Is the substitution:
$u=\sin^{-1}\cdot \frac{x}{3}$ or?
Because that would be strange.

**pickslides** · July 24th 2011, 02:21 PM

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

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

$\displaystyle x = 3\sin u$

So the expression becomes

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

Can you finish it from here?

Do you need to integrate this function? If so, consider $\displaystyle x = 3\sin u \implies dx = 3\cos u~du$

**JamesBond16** · July 24th 2011, 02:24 PM

yep thats it

**JamesBond16** · July 24th 2011, 02:25 PM

Originally Posted by pickslides

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

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

$\displaystyle x = 3\sin u$

So the expression becomes

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

Can you finish it from here?

huh im not sure.

**pickslides** · July 24th 2011, 02:29 PM

Originally Posted by JamesBond16

huh im not sure.

Give it a shot yourself, start by cleaning up the denominator.

Then take out any common factors inside the radical, then apply the identity $\displaystyle \sin^2 x +\cos^2 x = 1$ to eliminate the radical.

**JamesBond16** · July 24th 2011, 02:46 PM

[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?

**pickslides** · July 24th 2011, 02:55 PM

Nice work old mate!

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