Evaluate the integral using the indicated trigonometric substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^3}{\sqrt[\phantom{2}]{x^{2}36}}dx$;$x=6tan$

a$.(x^{2}36{)}^{\frac{3}{2}}-6\sqrt[\phantom{2}]{x^{2}36}C$

b$.\frac{1}{3}(x^{2}36{)}^{\frac{3}{2}}-36\sqrt[\phantom{2}]{x^{2}36}C$

c$.(x^{2}36{)}^{\frac{3}{2}}-\sqrt[\phantom{2}]{x^{2}36}c$

d$.\frac{1}{3}(x^{2}36{)}^{\frac{3}{2}}-\sqrt[\phantom{2}]{x^{2}36}c$

e$.\frac{3}{2}(x36{)}^{\frac{3}{2}}-36\sqrt[\phantom{2}]{x36}C$