∫ x^3 √(16-x^2) dx evaluated between [0, 4]
I know this is solved using trigonometric substitution, but I'm not sure how to work it. Please show steps so I will understand.
since cos^2θ = 1-sin^2θ, let
x = 4sinθ
16-x^2 = 16-16sin^2θ = 16cos^2θ
dx = 4cosθ dθ
Now you have an integrand of
(4sinθ)^3 (4cosθ) (4cosθ dθ)
= 1024 sin^3(θ) cos^2(θ) dθ
see what you can do with that