Question

Integrate the function $\frac{1}{\sqrt{8+3x-x^2}}$

Answer 1

$8+3x-x^2=8-(x^2-3x+\frac{9}{4}-\frac{9}{4})$
$=\frac{41}{4}-{(x-\frac{3}{2})}^2$
$\Rightarrow \int \frac{1}{\sqrt{8+3x-x^2}}dx=\int \frac{1}{\sqrt{\frac{41}{4}-{(x-\frac{3}{2})}^2}}dx$
Let $x-\frac{3}{2}=t$
$\therefore dx=dt$
$\Rightarrow \int \frac{1}{\sqrt{\frac{41}{4}-{(x-\frac{3}{2})}^2}}dx=\int \frac{1}{\sqrt{{\left(\frac{\sqrt{41}}{2}\right)}^2-t^2}}dt$
$={\sin }^{-1}\left(\frac{t}{\frac{\sqrt{41}}{2}}\right)+C$
$={\sin }^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+C$
$={\sin }^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C$