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$6\mathrm{.}3$ Let's Practice $1-$ Integrals Involving a Difference of Squares Part $1(5$ of $14)$

Score: $0/10$ Answered: $0/2$

$\sqrt[\phantom{2}]{0}$sub, $!$

Part a$)$

Use a trigonometric substitution to evaluate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{\sqrt[\phantom{2}]{9-x^2}}.($If necessary, use $C$ to represent a constant.$)$

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

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Part b$)$

For the substitution you used in Part a$),$ what are the restrictions on $?$ Use interval notation to enter your answer. For example, if the restrictions are we would enter $(-\frac{}{3},\frac{}{4}].$ If necessary, use a comma separated list to enter multiple intervals.

Restrictions on $$ : $$

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