# Thread: By making the substitution..

1. ## By making the substitution..

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

2. ## Re: By making the substitution..

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

3. ## Re: 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 $\displaystyle \displaystyle x = 3\sin u \implies dx = 3\cos u~du$

yep thats it

5. ## Re: By making the substitution..

Originally Posted by pickslides
$\displaystyle \displaystyle u=\sin^{-1} \frac{x}{3}$

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

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

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?
huh im not sure.

6. ## Re: By making the substitution..

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 $\displaystyle \displaystyle \sin^2 x +\cos^2 x = 1$to eliminate the radical.

7. ## Re: 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?

8. ## Re: By making the substitution..

Nice work old mate!