Answer
$${\sin ^{ - 1}}\left( {\frac{4}{5}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{1}{{\sqrt {25 - {x^2}} }}} dx \cr
& {\text{From the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ then}} \cr
& \int_0^4 {\frac{1}{{\sqrt {25 - {x^2}} }}} dx = \left[ {{{\sin }^{ - 1}}\left( {\frac{x}{5}} \right)} \right]_0^4 \cr
& {\text{Evaluating}} \cr
& = {\sin ^{ - 1}}\left( {\frac{4}{5}} \right) - {\sin ^{ - 1}}\left( {\frac{0}{5}} \right) \cr
& {\text{Simplifying}} \cr
& = {\sin ^{ - 1}}\left( {\frac{4}{5}} \right) \cr} $$
