Find the integral using an appropriate trigonometric substitution.
$$\int x \sqrt { 9 - x ^ { 2 } } d x$$
A) $\frac { 1 } { 3 } x ^ { 2 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$
B) $- \frac { 1 } { 3 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$
C) $- \frac { 1 } { 3 } x ^ { 2 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$
D) $\frac { 1 } { 3 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$

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The Answer of Find the integral using an appropriate trigonometric substitution.
$$\int x \sqrt { 9 - x ^ { 2 } } d x$$
A) $\frac { 1 } { 3 } x ^ { 2 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$
B) $- \frac { 1 } { 3 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$
C) $- \frac { 1 } { 3 } x ^ { 2 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$
D) $\frac { 1 } { 3 } \left( 9 - x ^ { 2 } \right) ^ { 3 / 2 } + C$

Find the Integral Using an Appropriate Trigonometric Substitution ∫x9−x2dx\int x \sqrt { 9 - x ^ { 2 } } d x∫x9−x2​dx

Find the Integral Using an Appropriate Trigonometric Substitution $\int x \sqrt { 9 - x ^ { 2 } } d x$