Question

Evaluate
${\int }_0^2\frac{1}{\sqrt{3+2x-x^2}}dx$

Answer 1

Given,

${\int }_0^2\frac{1}{\sqrt{3+2x-x^2}}dx$

$={\int }_0^2\frac{1}{\sqrt{-{(x-1)}^2+4}}dx$

substitute $u=x-1$ and $du=dx$

$={\int }_{-1}^1\frac{1}{\sqrt{-u^2+4}}du$

substitute $u=2\sin v$

$={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}1dv$

$={\left[v\right]}_{-\frac{\pi }{6}}^{\frac{\pi }{6}}$

$=\frac{\pi }{3}$