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Consider the following function. $âˆ«16âˆ’x_{2}â€‹x_{3}â€‹dx$ (a) Determine an appropriate trigonometric substitution. Use $x=4sin(Î¸)$, where $âˆ’2Ï€â€‹â‰¤Î¸â‰¤2Ï€â€‹$, since the integrand contains the expression $4_{2}âˆ’x_{2}â€‹$. Use $x=4tan(Î¸)$, where $âˆ’2Ï€â€‹<Î¸<2Ï€â€‹$, since the integrand contains the expression $4_{2}âˆ’x_{2}â€‹$. Use $x=4sec(Î¸)$, where $0â‰¤Î¸<2Ï€â€‹$ or $Ï€â‰¤Î¸<23Ï€â€‹$ since the integrand contains the expression $4_{2}âˆ’x_{2}â€‹$. (b) Apply the substitution to transform the integral into a trigonometric integral. Do not evaluate the integral. $âˆ«16âˆ’x_{2}â€‹x_{3}â€‹dx=âˆ«(Î¸$

Let , where . Then . Note that since , is positive.

Simplify terms.

Since is constant wi...

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