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What is the answer to this tricky little problem??

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A start: Express the thing inside the square root as $9-(x-3)^2$. Then make the substitution $x-3=3\sin\theta$.

Remark: We used the conventional completing the square approach, since that has to be learned. For this particular type of example, there is a nice shortcut.

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I do that, but I am not sure what to do after brain has gone dead... – Christopher Sep 14 '13 at 6:30
We get $dx=3\cos\theta\,d\theta$. Substitute. Note that $\sqrt{9-(x-3)^2}=\sqrt{9-9\sin^2\theta}=3\cos\theta$. So we need to find $\int 9\cos^2\theta\,d\theta$. You probably have already covered this integral. One way is to use the double-angle formula $\cos 2\theta=2\cos^2\theta -1$ to express $\cos^2\theta$ as $\frac{1+\cos 2\theta}{2}$. Now the integration is I hope easy. – André Nicolas Sep 14 '13 at 6:36
You can also do it by converting the thing inside the square to 9-(x-3)^2 and then applying the equation for integral of root of a^2-x^2. – Rajath Krishna R Sep 14 '13 at 6:41

Well, you can try doing this. Put $x= 6 \cdot \sin^{2}\theta$, then what basicall you have is the following \begin{align*} 6x-x^{2} &= 36\cdot \bigl(\sin^{2}\theta - \sin^{4}\theta\bigr) \end{align*}

Taking the square root you have the value as $6 \cdot \sin\theta \cdot \cos\theta$, so basically you want to solve the integral $$\int 3 \cdot \sin 2\theta \ d \theta$$

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