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

Question

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

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?

Answer 1

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..

Answer 2

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} $