Question

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

Answer 1

We have,

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

${\int }_0^1\frac{1}{\sqrt{3+1-1+2x-x^2}}dx{\int }_0^1\frac{1}{\sqrt{4-\left(1-2x+x^2\right)}}dx$

Since, $\left[\frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\frac{x}{a}\right]$

Therefore,

${\int }_0^1\frac{1}{\sqrt{{\left(2\right)}^2-{\left(x-1\right)}^2}}dx$

$I={\left[{\sin }^{-1}\frac{\left(x-1\right)}{2}\right]}_0^{1}I={\sin }^{-1}0-{\sin }^{-1}\left(-\frac{1}{2}\right)$

$I=-{\sin }^{-1}\frac{-1}{2}$

Hence, this is the answer.