Evaluate : `int_0^1 1/(sqrt(3+2x-x^2))dx`

Explanation

`int_0^1 1/(sqrt(3+2x-x^2))dx`

=`int_0^1 1/(sqrt(3(x^2)-2x-1)+1)dx`

`=int_0^1(1/sqrt((2)^2-(x-1)^2))dx`

`=[sin^(−1)(x−1)/2)]_0^1`

`=sin^(−1)(0)−sin^(−1)(−1/2)`

`=0−sin^(−1)(−sinπ/6)`

`=−sin^(−1)[sin(−π/6)]`

`=-(−π/6)=π/6.`