How To Integrate sqrt[x(4-x)] by Trigonometric Substitution

 

\begin{array}{l} \smallint \sqrt {x\left( {{\bf{4}} - x} \right)} \;\;{\rm{d}}x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;,Note\;\sqrt {a{x^2} + bx + c\;} \Rightarrow by\;Completing\;Square\\\\ \Rightarrow \smallint \sqrt { - \left( {{x^2} - 4x + 4 - 4} \right)} {\rm{d}}x\\ \Rightarrow \smallint \sqrt {4 - {{\left( {x - 2} \right)}^2}} {\rm{d}}x\;\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\;\;\;\;\;\;Let \Rightarrow 1 \cdot \left( {x - 2} \right) = 2\sin \theta \;\;\;\;\;\;\;\;\;\;\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{d}}x = 2\cos \theta {\rm{d}}\theta \\ Change{\rm{ }}the\;expression\\ \;\;\;\;\;\;{\left( {x - 2} \right)^2} = 4{\sin ^2}\theta \;\;\;\;\;\;\;\\ \;\;\;\;\;\;4 - {\left( {x - 2} \right)^2} = 4\left( {1 - {{\sin }^2}\theta } \right)\;\;\;\;\;\;\;\\ \;\;\;\;\;\;4 - {\left( {x - 2} \right)^2} = 4{\cos ^2}\theta \;\;\;\;\;\;\;\\ \;\;\;\;\;\sqrt {4 - {{\left( {x - 2} \right)}^2}} = 2\cos \theta \\ \smallint 2\cos \theta .2\cos \theta {\rm{d}}\theta = 4\smallint co{s^2}\theta \;d\theta \\ 4\smallint \frac{1}{2} + \frac{{\cos 2\theta \;}}{2}\;\;d\theta = 2\left( {\theta + \frac{{sin2\theta }}{2}} \right) + c\\ \Rightarrow 2\left( {\theta + \sin \theta \cos \theta } \right) + c \end{array}

\begin{array}{l} \Rightarrow 2{\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) + 2 \cdot \left( {\frac{{x - 2}}{2}} \right)\left( {\frac{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}{2}} \right) + c \end{array}