Q:

Evaluate $\int_{-2}^{-1}\frac{\text{d}x}{\sqrt{-x^2-6x}}$.

Problem statement [from Charlie Marshak's Math GRE Prep Problems]: 

Evaluate $\displaystyle \int\limits_{-2}^{-1}\dfrac{\text{d}x}{\sqrt{-x^2-6x}}$.

My work: notice that
$$\begin{align}
-x^2-6x &= -(x^2 - 6x) \\&= -\left[x^2-6x+\left(\dfrac{6}{2}\right)^2-\left(\dfrac{6}{2}\right)^2\right] \\
&= -(x^2-6x+9-9) \\
&= 9 - (x^2 - 6x+9) \\
&= 9 - (x-3)^2\text{.}
\end{align}$$
So,
$$\int\limits_{-2}^{-1}\dfrac{\text{d}x}{\sqrt{-x^2-6x}} = \int\limits_{-2}^{-1}\dfrac{\text{d}x}{3\sqrt{1-\left(\dfrac{x-3}{3}\right)^2}}\text{.}$$
If $u = \dfrac{x-3}{3}$, $\text{d}u = \dfrac{1}{3}\text{ d}x$, so that 
$$\int\limits_{-2}^{-1}\dfrac{\text{d}x}{3\sqrt{1-\left(\dfrac{x-3}{3}\right)^2}} = \int\limits_{-5/3}^{-4/3}\dfrac{\text{d}u}{\sqrt{1^2-u^2}} = \sin^{-1}\left(\dfrac{-4}{3}\right) - \sin^{-1}\left(\dfrac{-5}{3}\right)\text{.}$$
The answer is supposed to be $\sin^{-1}(2/3)-\sin^{-1}(1/3)$. Where did I go wrong?

A:

Where you have $x-3$ you should have $x+3$. The exact point you went wrong is in the first line, where you put $-6x$ rather than $+6x$. There is already a minus sign. This is the error:
$$
-x^2-6x \neq -(x^2 - 6x)
$$
$$
-x^2-6x = -(x^2 + 6x)
$$
Note also that $\sin^{-1}(x)$ is meaningless if $|x|>1$

