Evaluate the integral: ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{49-x^2}}{10x^2}dx$

$($A$)$ Which trig substitution is correct for this integral?

$x=49sin()$

$x=7sin()$

$x=7sec()$

$x=49tan()$

$x=49sec()$

$x=7tan()$

$($B$)$ Which integral do you obtain after substituting for $x$ and simplifying?

Note: to enter $,$ type the word theta.

J

$d$

$($C$)$ What is the value of the above integral in terms of $?$

$$

$+C$

$($D$)$ What is the value of the original integral in terms of $x?$

$\frac{1}{10}(-arcsin(\frac{1}{7}x)-\frac{\sqrt[\phantom{2}]{49-x^2}}{x})$