S=integral sign
S3x(4-x^2)^1/2
use trig substitution
I am confused
I got
x sin(sin^-1(x/2))+cos(sin^-1(x/2))
Because sin^2θ = 1-cos^2θ, the logical choice is
x = 2sinθ
4-x^2 = 4-4sin^2θ = 4cos^2θ
√(4-x^2) = 2cosθ
dx = 2cosθ dθ
Now your integral becomes
∫3x√(4-x^2) dx
= ∫3*2sinθ * 2cosθ 2cosθ dθ
= 24∫cos^2θ sinθ dθ
Now let u = cosθ and the integral is just
= -24∫u^2 du
= -24(1/3 u^3)
= -8u^3
= -8cos^3θ
= -(2cosθ)^3
= -(4-x^2)^(3/2) + C