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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.

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  • $\begingroup$ you don't even need any substitution; $\int z \sqrt{9-z^2}\, dz = -\frac{1}{3}(9-z^2)^{3/2}+C$ $\endgroup$ – abel Feb 7 '15 at 18:20
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$$I=\int z\sqrt{9-z^2}{\rm d}z$$ 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?

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Why trigonometry for such an easy problem ? Set $u=9-z^2$, then $du=-2zdz$ and thus, $$...=-\frac{1}{2}\int^{9-z^2}\sqrt{u}du=-\frac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}}\right]^{9-z^2}=...$$

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  • $\begingroup$ have you read my ps in the body of the question? $\endgroup$ – Guerlando OCs Feb 8 '15 at 1:52
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Since your question seeks to express $\cos \theta$ as a function of $z$, use a right-triangle, with theta the degree of the angle measured with respect to the positive x-axis. Then we know

$$z = 3\sin \theta \iff \frac z3 = \sin \theta$$ where $3$ is the hypotenuse of the triangle, and $z$ one of the legs (opposite $\theta$).

Then the other leg (adjacent to $\theta$) is given by $\sqrt{9 - z^2}$. And $$\cos \theta = \frac {\sqrt{9 - z^2}}3$$

Then the back substitution becomes $$-27\left(\frac{\cos^3\theta}{3}\right) +C= -9\left(\frac{\sqrt{9-z^2}}3\right)^{3} + C= -\frac 13(9-z^2)^{3/2}+C$$

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