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I am getting a sign error when evaluating:

$$ \int \dfrac {1} {\sqrt{-x^{2} - 4x}}dx$$

I completed the square in the denominator leaving me:

$$\int \dfrac {1} {\sqrt{-x^{2} - 4x + 4 - 4}}dx$$

$$\int \dfrac {1} {\sqrt{-(x^{2} + 4x - 4 + 4)}}dx$$

$$\int \dfrac {1} {\sqrt{-(x+2)^{2} +4}}dx$$

I then let $ u = x+2 , du = dx$, and $a = 2.$

$$\int \dfrac {du} {\sqrt{-u^{2} + a^{2}}}$$

$$\arcsin \dfrac {-(x+2)} {2} + C$$

However, the correct answer should be $$\arcsin \dfrac {x+2} {2} + C$$

Where did I go astray?

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$\int \dfrac {du} {\sqrt{-u^{2} + a^{2}}}=\arcsin\dfrac{u}{a}+C$ –  user21436 Apr 7 '12 at 21:37
    
Ah, wow. Thanks for the catch! –  Joe Apr 7 '12 at 21:39
    
Thanks for the edit Brian. I was using \arcsin before someone told me to use \operatorname arcsin at one point. I always preferred \arcsin. Is there a general consensus of what to use here? –  Joe Apr 7 '12 at 21:40
    
The function is predefined in the TeX library, so, just \arcsin looks good. For instance if you define a new function called Good, TeX won't know it, and it will look good with \operatorname{Good}$, you may also want to note that I use curly braces there--{}. –  user21436 Apr 7 '12 at 21:45
    
Cool, nice to know. Thanks. –  Joe Apr 7 '12 at 21:46
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2 Answers

up vote 2 down vote accepted

$\displaystyle \int \dfrac{du}{\sqrt{a^2 - u^2}} = \arcsin \dfrac{u}{a} + C$

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Ah, it's fine. You CW'ed it anyways. Thanks. –  Joe Apr 7 '12 at 21:41
    
I don't think you should make it CW. This solves the problem that OP has been interested in. My +1 for this. Please flag this for a mod and unwikify it, please. –  user21436 Apr 7 '12 at 21:42
    
(The C-Wikification looks as though it was part of the answer, but was a quick edit once I realized that you had already received an answer in comments!) –  The Chaz 2.0 Apr 7 '12 at 21:43
    
@TheChaz Only a mod can Unwikify it--you cannot roll-back that as if it was an edit. Note that a question can be made CW with only a mod's help while an answer you write can be made CW by yourself. I am flagging it for mod attention BTW. –  user21436 Apr 7 '12 at 21:48
    
I've removed the CW. –  Zev Chonoles Apr 7 '12 at 22:24
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\begin{align*} \int\frac{1}{\sqrt{-x^{2}-4x}} &=\int\frac{1}{\sqrt{-x^{2}-4x+4-4}}\ &=\int\frac{1}{\sqrt{2^{2}-(x+2)^{2}}}\ &=\sin^{-1}(\frac{x+2}{2})+c. \end{align*} Here i use $\int\frac{1}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}(\frac{x}{a})+c.$

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