Question

Integrate the function $\frac{1}{\sqrt{7-6x-x^2}}$

Answer 1

$7-6x-x^2=7-(x^2+6x+9-9)=7-(x^2+6x+9-9)$
$=16-(x^2+6x+9)=16-(x+3{)}^2=(4{)}^2-(x+3{)}^2$
$\therefore \int \frac{1}{\sqrt{7-6x-x^2}}dx=\int \frac{1}{\sqrt{(4{)}^2-(x+3{)}^2}}dx$
Let $x+3=t\Rightarrow dx=dt$
$\Rightarrow \int \frac{1}{\sqrt{(4{)}^2-(x+3{)}^2}}dx=\int \frac{1}{\sqrt{(4{)}^2-(t{)}^2}}dt$
$={\sin }^{-1}\left(\frac{t}{4}\right)+C$
$={\sin }^{-1}\left(\frac{x+3}{4}\right)+C$