Math  Trig Substitution
posted by Marissa .
How can I solve the integral of x^3√(9x^2) dx using trigonometric substitution? ?
∫ x^3√(9x^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√(9x^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 √ (1sin^2)θ
1  sin^2 = cos^2
= ∫ (27sin^3θ * (3  3 COS θ )* 3cosθdθ