# Integrate $\frac{1}{\sqrt{6x-x^2-5}}$

Question: Integrate $\frac{1}{\sqrt{6x-x^2-5}}$

My Working:

$$\int{\frac{1}{\sqrt{6x-x^2-5}}} = \int{\frac{1}{\sqrt{-(x^2-6x+5)}}} = \int{\frac{1}{\sqrt{-((x-3)^2-3^2+5)}}} = \int{\frac{1}{\sqrt{-((x-3)^2-4)}}}$$

Is it right so far? Because I think I should be using the formula

$$\int{\frac{1}{\sqrt{a^2-(x+b)^2}}} = \sin^{-1} \left( \frac{x+b}{a} \right)+c$$

$-4$ can't fit into $a^2$ can it? I haven't covered complex numbers yet, so I must have made a mistake?

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1
There's two minus signs in front of the 4. You have $a^2=4$ and $b=-3$ by my reckoning. – anon Oct 17 '11 at 7:16
$6x-x^2-5=2^2-(x-3)^2$ – pedja Oct 17 '11 at 7:20

## 2 Answers

Your derivation is correct, but you aren't looking at the last line correctly. Notice that $$-((x-3)^2-4)=4-(x-3)^2,$$ so your parameters are $a=2$ and $b=-3$ to plug into the formula. We rule out $a=-2$ because the square root function is always nonnegative, and so following holds only for $a$ positive: $$\frac{d}{dx}\sin^{-1}\left(\frac{x+b}{a}\right)=\frac{1}{a\sqrt{1-(x+b)^2/a^2}}=\frac{1}{\sqrt{a^2-(x+b)^2}}.$$

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You are doing just fine. You completed the square, which is the key step, and ended up with $$\int \frac{dx}{\sqrt{4-(x-3)^2}}.$$

Now you have a formula that you want to use. It may be better not to try to remember too many formulas. We will instead find a substitution that gives us a very familiar integral.

The above integral looks kind of like $$\int \frac{dw}{\sqrt{1-w^2}}.$$ We look for a substitution that brings out the resemblance. Suppose that we let $x-3=2u$. Then the bottom will look like $\sqrt{4-4u^2}$, which is good, because we can then "take the $4$'s out."

So let's do it. Let $x-3=2u$, or if you prefer, let $(x-3)/2=u$. Then $dx=2\,du$ and our integral becomes $$\int\frac{2u}{\sqrt{4-4u^2}}.$$

But $\sqrt{4-4u^2}=2\sqrt{1-u^2}$. So our integral simplifies to $$\int \frac{du}{\sqrt{1-u^2}}=\arcsin u +C.$$ Finally, replace $u$ by $(x-3)/2$.

Another way to get to the right substitution is to note that $$\sqrt{4-(x-3)^2}=2\sqrt{1-((x-3)/2)^2},$$ which makes letting $u=(x-3)/2$ very natural.

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