How To Integrate sqrt[-x^2+2x+5] by Trigonometric Substitution

 

\begin{array}{l} \smallint \sqrt { - \left( {{x^{\bf{2}}} - {\bf{2}}x - {\bf{5}}} \right)} \;{\rm{d}}x\;\;\;\;\;\;\;\;\;\;,Note\;\sqrt {a{x^2} + bx + c\;} \Rightarrow by\;Completing\;Square\\\\ \Rightarrow \smallint \sqrt { - \left( {{x^2} - 2x + 1 - 1 - 5} \right)} \;{\rm{d}}x\\ \Rightarrow \smallint \sqrt {6 - {{\left( {x - 1} \right)}^2}} \;{\rm{d}}x\;\;\;\;\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\;\;\;Let \Rightarrow x - 1 = \sqrt 6 \sin \theta \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{d}}x = \sqrt 6 \cos \theta d\theta \\ Change{\rm{ }}the\;expression\\ \;\;\;\;\;\;\;6 - {\left( {x - 1} \right)^2} = 6\left( {1 - {{\sin }^2}\theta } \right)\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\;6 - {\left( {x - 1} \right)^2} = 6{\cos ^2}\theta \;\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\sqrt {6 - {{\left( {x - 1} \right)}^2}} = \sqrt 6 \cos \theta \\ \smallint \sqrt 6 \cos \theta \sqrt 6 \cos \theta {\rm{d}}\theta = 6\smallint co{s^2}\theta d\theta \\ 6\smallint \frac{{1 + \cos 2\theta }}{2}{\rm{d}}\theta = 3\left( {\theta + \frac{{sin2\theta }}{2}} \right) + c\\ \Rightarrow 3\theta + 3\;\sin \theta \cos \theta + c \end{array}

\begin{array}{l} \Rightarrow 3{\sin ^{ - 1}}(\frac{{x - 1}}{{\sqrt 6 }}) + 3 \cdot \left( {\frac{{x - 1}}{{\sqrt 6 }}} \right) \cdot \left( {\frac{{\sqrt {6 - {{\left( {x - 1} \right)}^2}} }}{{\sqrt 6 }}} \right) + c \end{array}