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

∫ 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?

∫ 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

To further simplify the integral before solving it, you can use trigonometric identities to express powers of sine in terms of cosine.

Recall that sin^2θ + cos^2θ = 1. Rearranging this equation, we have sin^2θ = 1 - cos^2θ.

Substitute this identity into the integral:

∫ (27sin^3θ * (3 - 3sinθ)* 3cosθdθ
= ∫ (27sin^3θ * (3 - 3(1 - cos^2θ))* 3cosθdθ
= ∫ (27sin^3θ * (0 + 3cos^2θ))* 3cosθdθ
= ∫ (27sin^3θ * 3cos^2θ) * 3cosθdθ.

Now, simplify the expression further:

∫ (81sin^3θ * cos^2θ) * cosθdθ
= 81∫ sin^3θ * cos^3θdθ.

We can now solve this integral using the trigonometric identity for the product of powers of sine and cosine:

∫ sin^mθ * cos^nθdθ = (1/2) * (m-1) * (n+1) * ∫ sin^(m-2)θ dθ,

where m and n are positive integers.

In this case, m = 3 and n = 3, so:

∫ sin^3θ * cos^3θdθ
= (1/2) * (3-1) * (3+1) * ∫ sin^(3-2)θ dθ
= 2 * 4 * ∫ sinθ dθ
= 8 ∫ sinθ dθ.

Now, integrate ∫ sinθ dθ:

∫ sinθ dθ = -cosθ + C,

where C is the constant of integration.

Finally, substitute back the original variable:

8(-cosθ) + C = -8cosθ + C.

Therefore, the solution to the integral of x^3√(9-x^2) dx using trigonometric substitution is:

-8cosθ + C.

Remember to convert the answer back into terms of x by substituting θ with its original expression in terms of x, which was x = 3sinθ.