solving $\int \frac{dx}{\sqrt{-x^2-12x+28}}$ 
$$\int \frac{dx}{\sqrt{-x^2-12x+28}}$$

First we need to use completing the square $-(x^2+12x-28)=-(x+6)^2+64$
So we have $\int \frac{dx}{\sqrt{-(x+6)^2+64}}$ I know that it is a general form of $\arcsin(\frac{x+6}{8})$ but how can I solve it using substitution?  

Its not really necessary to go for substitution. Use basic differentiation of inverse trig functions.
$\begin{align}
\int\frac{dx}{\sqrt{a^2-x^2}}&=
\sin^{-1}(\frac{x}{a})+C\\
&=\int\frac{dx}{\sqrt{64-(x+6)^2}}\\
&=\sin^{-1}\frac{(x+6)}{8}+C
\end{align}
$
