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.