Evaluate the following integral using trigonometric substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{\sqrt[\phantom{2}]{x^2-49}},x>7$

$$ A$.x=7tan$

B$.x=7sin$

C$.x=7sec$

Rewrite the given integral using this substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{\sqrt[\phantom{2}]{x^2-49}}={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}},d$

$($Type an exact answer.$)$

Evaluate the integral.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{\sqrt[\phantom{2}]{x^2-49}}=$