Question

Integrate the following functions.
$\int \frac{1}{\sqrt{8+3x-x^2}}dx$

Answer 1

Let $I=\int \frac{1}{\sqrt{8+3x-x^2}}dx=\int \frac{1}{\sqrt{8-\sqrt{[x^2-3x+(\frac{3}{2}{)}^2-\left(\frac{3}{2}{)}^2\right]}}}dx$
$=\int \frac{1}{\sqrt{8-\left[\right(x-\frac{3}{2}{)}^2-\frac{9}{4}]}}dx=\int \frac{1}{\sqrt{8+\frac{9}{4}-(x-\frac{3}{2}{)}^2}}dx=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2}{)}^2-(x-\frac{3}{2}{)}^2}}dx$

Let $x-\frac{3}{2}=t\Rightarrow dx=dt$
$\therefore I=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2}{)}^2-t^2}}dt=si{n}^{-1}\left(\frac{t}{\frac{\sqrt{41}}{2}}\right)+C[\because \int \frac{dx}{\sqrt{a^2-x^2}}=si{n}^{-1}(\frac{x}{a}\left)\right]=si{n}^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+C=si{n}^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C(\because t=x-\frac{3}{2})$