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How can i evaluate this indefinite integral.

$$ \int { \frac { 1 }{ \sqrt { 8-{ x }^{ 2 }-2x } } } \,dx$$

I know it involves completing square but i don't know how to do it.

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    $\begingroup$ hint: $8-x^2-2x=8-(x^2+2x)=3^2-(x+1)^2$ $\endgroup$ – oldrinb Jan 12 '14 at 21:08
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Hint:

$$8 - x^2 - 2x = 8 - (x^2 + 2x) = 8 - (x^2 + 2x + 1) + 1 = 9 - (x + 1)^2$$

Can you take it from here?

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  • $\begingroup$ Why did you add the +1 after the parentheses. Why not -1 ? $\endgroup$ – Out Of Bounds Jan 12 '14 at 21:17
  • $\begingroup$ Distribute the minus sign before the term in parentheses. $\endgroup$ – user61527 Jan 12 '14 at 21:19
  • $\begingroup$ Ohh yeah got it. Thank you :) $\endgroup$ – Out Of Bounds Jan 12 '14 at 21:20
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Hints:

$$8-x^2-2x=9-(x+1)^2\implies \sqrt{8-x^2-2x}=3\sqrt{1-\left(\frac{x+1}3\right)^2}$$

and now do remember the derivative of $\;\arcsin\;$ ...

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