# 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?

• try $\sin \theta = \frac{x+6}{8}$ – vnd Jan 4 '16 at 18:45

Put $y=\frac{x+6}{8}$ then the integral is equivalent to $$\int \frac{\mathrm{d}y}{\sqrt{1-y^2}}.$$ From here you can conclude..
\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}