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Evaluate the following definite integral.

$ \int _{ 3 }^{ 6 }{ \frac { 1 }{ \sqrt { 27+6x-{ x }^{ 2 } } } } \quad dx $

I can't figure out how to complete the square.

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  • $\begingroup$ $ 27+(6x-x^2) $ ? $\endgroup$ – Out Of Bounds Jan 13 '14 at 1:50
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    $\begingroup$ Perhaps write $$27 - (x^2 - 6x)$$ instead. How would you deal with just $x^2 - 6x$? $\endgroup$ – user61527 Jan 13 '14 at 1:51
  • $\begingroup$ Also, on a different note: Please use titles that describe your problem; your last three questions have identical vague titles, and so do the previous three. $\endgroup$ – user61527 Jan 13 '14 at 1:58
  • $\begingroup$ I got it. Thank you :) $\endgroup$ – Out Of Bounds Jan 13 '14 at 2:07
  • $\begingroup$ Can you give me an example of title that describe the problem well for this one ? $\endgroup$ – Out Of Bounds Jan 13 '14 at 2:08
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Write the relevant term as

$$27 - (x^2 - 6x)$$

The usual procedure for $x^2 + bx$ is to add $(b/2)^2$; in this case, $b = 6$ and we add $9$. This leads us to \begin{align*} 27 - (x^2 - 6x) &= 27 - (x^2 - 6x + 9) + 9 \\ &= 36 - (x - 3)^2 \end{align*}

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Hint:

Write

$$\frac { 1 }{ \sqrt { 27+6x-{ x }^{ 2 } } } = \frac{1}{6\sqrt{1-\frac{\left(x-3\right)^2}{36}} }$$

And subsitute $$x = 3 + u$$

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