Evaluate the following integral.

${\displaystyle \underset{0}{\overset{\frac{5}{2}}{}}}\frac{dx}{\sqrt[\phantom{2}]{25-x^2}}$

Which substitution transforms the given integral into one that can be evaluated directly in terms of $?$

A$.x=5tan$

B$.x=5sin$

C$.x=5sec$

For this substitution, $dx=()d0.$

$$

${\displaystyle \underset{0}{\overset{\frac{5}{2}}{}}}\frac{dx}{\sqrt[\phantom{2}]{25-x^2}}=$

$$

$($Type an exact answer, using $$ as needed.$)$