Evaluate the following integral using trigonometric substitution.

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

A$.x=5sin$

B$.x=5sec$

C$.x=5tan$

Rewrite the given integral using this substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{25-x^2}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}},d0$

$($Type an exact answer.$)$

Evaluate the integral.

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

$($Type an exact answer.$)$