# Integral Of $\sqrt{9-y^2}dy$

I would like to integrate the following:
$$\int\sqrt{9-y^2}dy$$ What I did:

1. $y=3\sin t$
2. $dy=3\cos tdt$

$$\int\sqrt{9-9\sin ^2 t}\cdot 3\cos t dt=\int\sqrt{9}\cdot\sqrt{1-\sin^2 t}\cdot 3\cos tdt=\int 9\cos ^2tdt$$ $$9\cdot \int(\frac{1}{2}+\frac{\cos 2t}{2}) dt=9\cdot\left [ t+\frac{\sin 2t}{2}\right]$$ I'm pretty stuck now, I would like to get some advice
How its suppose to be now?
$$\dots=9\left[ \arcsin(\frac{y}{3})+\frac{\sin( 2 \cdot \arcsin(\frac{y}{3}))}{2}\right]?$$

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Note that $$\sin 2t = 2 \sin t \cos t$$

Now use the fact that $\sin^2 t + \cos^2 t = 1$ to rewrite $\cos t$ in terms of $y$. This leads to

\begin{align*} \frac{\sin 2t}{2} &= \sin t \cos t \\ &= \sin t \sqrt{1 - \sin^2 t} \\ &= \frac{y}{3} \sqrt{1 - \left(\frac{y}{3}\right)^2} \\ &= \frac{1}{9} y \sqrt{9 - y^2} \end{align*}

You've got a very small mistake, though, in your application of the double angle formula: You should have $$\frac 1 2 + \frac{\cos 2t}{2}$$

Fixing this and applying the above leads to the simplified solution from W|A.

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Sorry, I did not understand you, there is something wrong in my way? I'll do it all over again if needed. Where I can use it? – Ofir Attia Jan 1 '14 at 20:03
@OfirAttia I've expanded my answer. – user61527 Jan 1 '14 at 20:08