So do the subs.
x = 3sinθ
dx = 3cosθ dθ
√(9-x^2) = √(9-9sin^2θ) = 3cosθ
so now you have
∫5x^3 √(9−x^2) dx
= ∫5 (3sinθ)^3 * 3cosθ * 3cosθ dθ
= 5*3^5 ∫ sin^3θ cos^2θ dθ
= 5*3^5 ∫ sin^3θ - sin^5θ dθ
= 5*3^5 * 1/30 cos^3θ (3cos2θ - 7) + C
= 5*3^5 * 1/30 cos^3θ * 2/3 (9sin^2θ + 6) + C
= -(x^2+6) (9-x^2)^(3/2) + C
Evaluate the integral using the indicated trigonometric substitution.
∫5x^3(sqrt(9−x^2))dx and x =3sin(θ)
2 answers
their suggestion:
let x = 3sinθ
then x^2 = 9sin^2 θ
and x^3 = 27sin^3 θ
dx = 3cosθ dθ
∫5x^3(√(9−x^2))dx
= 5 ∫ (27sin^3 θ)√(9−9sin^2 θ) 3cosθ dθ
= 5 ∫ (27sin^3 θ)(3)√(1- sin^2 θ) 3cosθ dθ
= 1215 ∫ (sin^3 θ)(cosθ)(cosθ) dθ
= 1215 ∫ (sin^3 θ)(cos^2 θ) dθ
sort of stuck here.
wolfram shows this:
www.wolframalpha.com/input/?i=%E2%88%AB%28%285x%5E3%29%E2%88%9A%289-x%5E2%29+%29dx
let x = 3sinθ
then x^2 = 9sin^2 θ
and x^3 = 27sin^3 θ
dx = 3cosθ dθ
∫5x^3(√(9−x^2))dx
= 5 ∫ (27sin^3 θ)√(9−9sin^2 θ) 3cosθ dθ
= 5 ∫ (27sin^3 θ)(3)√(1- sin^2 θ) 3cosθ dθ
= 1215 ∫ (sin^3 θ)(cosθ)(cosθ) dθ
= 1215 ∫ (sin^3 θ)(cos^2 θ) dθ
sort of stuck here.
wolfram shows this:
www.wolframalpha.com/input/?i=%E2%88%AB%28%285x%5E3%29%E2%88%9A%289-x%5E2%29+%29dx