urgent!! help please

Question

integral 0 to sqrt 3 of x^3/sqrt (4-x^2) dx

please show all steps to evaluate this integral

Answer 1

∫_{(0 to √3)} [x³ / √(4 - x²)]dx

Let x = 2sinθ, then dx = 2cosθdθ.

When x = 0, θ = Sin^-1(0/2) = 0 and when x = √3, θ = Sin^-1(√3/2) = π/3.

So we have ∫_{(0 to π/3)} [( 8sin³θ / √(4 - 4sin²θ))(2cosθ)]dθ

= 8 ∫_{(0 to π/3)} sin³θ dθ = 8 ∫_{(0 to π/3)} sin²θ sinθ dθ = 8∫_{(0 to π/3)} (1 - cos²θ) sinθ dθ

Now, let u = cosθ, then du = -sinθdθ

When θ = 0, u = cos0 = 1 and when θ = π/3, u = cos(π/3) = 1/2

So, we have -8∫_{(1 to 1/2)} (1 - u²)du = 8 ∫_{(1/2 to 1)} (1 - u²) du

= ( 8u - (8/3)u³)_{(1/2 to 1)} = (8 - 8/3) - (4 - 1/3) = 4 - 7/3 = 5/3

Answer 2