Question

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

Answer 1

To find $\int \frac{1}{\sqrt{8+3x-x^2}}.dx$

$=\int \frac{1}{\sqrt{8-(x^2-3x)}}.dx$

$=\int \frac{1}{\sqrt{8-[x^2-3x+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2]}}.dx$

$=\int \frac{1}{\sqrt{8-[{(x-\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2]}}.dx$

$=\int \frac{1}{\sqrt{8-{(x-\frac{3}{2})}^2+\frac{9}{4}}}.dx$

$=\int \frac{1}{\sqrt{8+\frac{9}{4}-{(x-\frac{3}{2})}^2}}.dx$

$=\int \frac{1}{\sqrt{\frac{41}{4}-{(x-\frac{3}{2})}^2}}.dx$

$=\int \frac{1}{\sqrt{{\left(\frac{\sqrt{41}}{4}\right)}^2-{(x-\frac{3}{2})}^2}}.dx$

$={\sin }^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+C$

$[\because \int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+C]$

$={\sin }^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C$

where $C$ is a constant of integration.