Evaluate the following integrals:
$\int \frac{1}{\sqrt{8+3 x-x^{2}}} d x$
$8+3 x-x 2$ can be written as $8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$
Therefore
$8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$
$=\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}$
$\int \frac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x$
Let $x-3 / 2=t$
$d x=d t$
$\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x=\int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^{2}-t^{2}}} d t$
$=\sin ^{-1}\left(\frac{\mathrm{t}}{\frac{\sqrt{41}}{2}}\right)+\mathrm{c}$
$\left[\right.$ since $\left.\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]$
$=\sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+c$
$=\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+c$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.