Evaluate:: ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{6,400-x^2}}dx$

$($i$)$ What is the correct trigonometric substitution to find an antiderivative:

$x=6,400sec()$

$x=80sin()$

$x=80tan()$

$x=80sec()$

$x=80sec()tan()$

$($ii$)$ Which one is the correct statement after substitting the above trigonometric function

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{6,400-x^2}}dx=\frac{1}{{80}^2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{s}^2()d$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{6,400-x^2}}dx=\frac{1}{{80}^2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}se{c}^2()d$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{6,400-x^2}}dx=\frac{1}{{80}^2}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}cs{c}^2()d$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{6,400-x^2}}dx=\frac{1}{80}{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}ta{n}^2()d$

$($iii$)$ What is the most general antiderivative of $\frac{1}{x^2\sqrt[\phantom{2}]{6,400-x^2}}$

Answer:

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{-x^2\sqrt[\phantom{2}]{6,400-x^2}}dx=$