calculus - Integrate $int zsqrt{9-z^2}dz$ by trigonometric subsitution

I want to integrate $\int z\sqrt{9-z^2}$ by trigonometric subsitution. Here's what I did:

let $z = 3\sin\theta\implies dz = 3\cos\theta d\theta$

$$\int z\sqrt{9-z^2}dz = \int 3\sin \theta\sqrt{9-3^2\sin^2\theta} \ \ 3\cos\theta \ \ d\theta = 9\int \sin\theta\cos\theta\sqrt{9(1-\sin^2\theta)} \ \ d\theta = \\9\int\sin\theta\cos\theta\ 3|\cos\theta| d\theta = 27\int\sin\theta\cos^2\theta \ d\theta$$
let $u = \cos\theta\implies du = \sin\theta d\theta$

$$27\int\sin\theta\cos^2\theta \ d\theta = -27\int u^2\ du = -27\frac{u^3}{3} = -27\frac{\cos^3\theta}{3}$$
Since $z=3\sin\theta$ then $z^2 = 9\sin^2\theta\implies z^2 = 9(1-\cos^2\theta)\implies z^2 = 9-9\cos^2\theta \implies 9\cos^2\theta = 9-z^2\\\implies \cos^2\theta = \dfrac{9-z^2}{9}$

But how to find $\cos^3(\theta)$ in terms of $z$?

ps: I know this integral is a lot easier to integrate with simple substitution.

Answer

$$I=\int z\sqrt{9-z^2}{\rm d}z$$
Let $z=3\sin\theta$:
$$I=\int 3\sin\theta.3\cos\theta3\cos\theta.{\rm d}\theta\\=27\int\cos^2\theta\sin\theta{\rm d}\theta\\=-27\int\cos^2\theta{\rm d}(\cos \theta)\\=-27\frac{\cos^3\theta}3\\=-9(1-(z/3)^2)^{3/2}+c$$
Because $\sin^2\theta+\cos^2\theta=1\implies \cos\theta=\sqrt{1-\sin^2\theta}=(1-(z/3)^2)^{1/2}$
Hope you can cube it now?