In Exercises $5-16,$ apply Trigonometric Substitution to evaluate the indefinite integrals.

$5.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{x^{2}1}dx$

$6.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{x^{2}4}dx$

$7.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{1-x^2}dx$

$8.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{9-x^2}dx$

$9.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{x^2-1}dx$

$10.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{x^2-16}dx$

$11.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{4x^{2}1}dx$

$12.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{1-9x^2}dx$

$13.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{16x^2-1}dx$

$14.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{8}{\sqrt[\phantom{2}]{x^{2}2}}dx$