Answer
$$\frac{{25}}{4}\arcsin \left( {\frac{{2x}}{5}} \right) + \frac{1}{2}x\sqrt {25 - 4{x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {25 - 4{x^2}} dx} \cr
& {\text{Rewrite}} \cr
& \int {\sqrt {25 - 4{x^2}} dx} = \int {\sqrt {{{\left( 5 \right)}^2} - {{\left( {2x} \right)}^2}} dx} \cr
& {\text{Let }}u = 2x,{\text{ and }}a = 5,{\text{ }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& {\text{Substitute}} \cr
& \int {\sqrt {{{\left( 5 \right)}^2} - {{\left( {2x} \right)}^2}} dx} = \int {\sqrt {{a^2} - {u^2}} \left( {\frac{1}{2}} \right)d} u \cr
& = \frac{1}{2}\int {\sqrt {{a^2} - {u^2}} d} u \cr
& {\text{Use the special integration formula }}\left( {{\text{Theorem 8}}{\text{.2}}} \right) \cr
& \int {\sqrt {{a^2} - {u^2}} du = \frac{1}{2}\left( {{a^2}\arcsin \frac{u}{a} + u\sqrt {{a^2} - {u^2}} } \right) + C} \cr
& \frac{1}{2}\int {\sqrt {{a^2} - {u^2}} d} uu = \frac{1}{4}\left( {{a^2}\arcsin \frac{u}{a} + u\sqrt {{a^2} - {u^2}} } \right) + C \cr
& {\text{Substitute back}} \cr
& = \frac{1}{4}\left( {{{\left( 5 \right)}^2}\arcsin \left( {\frac{{2x}}{5}} \right) + 2x\sqrt {{{\left( 5 \right)}^2} - {{\left( {2x} \right)}^2}} } \right) + C \cr
& = \frac{{25}}{4}\arcsin \left( {\frac{{2x}}{5}} \right) + \frac{1}{2}x\sqrt {25 - 4{x^2}} + C \cr} $$
