How can I solve the integral of x^3√(9-x^2) dx using trigonometric substitution? ?

Question

∫ x^3√(9-x^2) dx

So then I know that
x = 3sinθ
dx = 3cosθdθ

When I substitute, it becomes

∫ (3sinθ)^3 * √(9-(3sinθ)^2) * 3cosθdθ

= ∫ (27sin^3θ * (3 - 3sinθ)* 3cosθdθ

Is there any way to furter simplify this before I solve it? And if there isn't, how would I go about solving it?

Answer 1

∫ x^3√(9-x^2) dx

So then I know that
x = 3sinθ
dx = 3cosθdθ

When I substitute, it becomes

∫ (3sinθ)^3 * √(9-(3sinθ)^2) * 3cosθdθ

BUT √(9-(3sinθ)^2) = 3 √ (1-sin^2)θ
1 - sin^2 = cos^2

= ∫ (27sin^3θ * (3 - 3 COS θ )* 3cosθdθ