Answer: [tex]\bold{\dfrac{5}{3}}[/tex]
Step-by-step explanation:
First, simplify 2 sin(π/3) = 2 (√3/2) = √3
So the integral is from x = 0 to x = √3
FYI: I will be using double substitution.
[tex]\int\limits^{\sqrt3}_0 {\dfrac{x^3}{\sqrt{4-x^2}}} \, dx \quad \longrightarrow \quad \int\limits^{\sqrt3}_0 {\dfrac{x\cdot x^2}{\sqrt{4-x^2}}} \, dx\\\\\\\\\text{Let v}=x^2\quad \text{then dv}=2x\ dx\\\\ \int\limits^{\sqrt3}_0 \bigg(\dfrac{1}{2}\bigg){\dfrac{v}{\sqrt{4-v}}} \, dv\\\\\\\\\text{Let u}=4-v\quad \text{then du}=-1\ dv\\.\ \rightarrow \text{v}=4-u\quad\ \rightarrow \text{dv}=-1\ du\\\\ \int\limits^{\sqrt3}_0 \bigg(-\dfrac{1}{2}\bigg){\dfrac{4-u}{\sqrt{u}}} \, du\\[/tex]
[tex]= -\dfrac{1}{2}\ \int\limits^{\sqrt3}_0 \bigg({\dfrac{4}{\sqrt{u}} -\dfrac{u}{\sqrt{u}}}\bigg)\, du\\\\\\= -\dfrac{1}{2}\ \int\limits^{\sqrt3}_0 \bigg({4(u)^{-\dfrac{1}{2}}}-{(u)^{\dfrac{1}{2}}}\bigg)\ du\\\\\\\\\text{Integrate:}\\\\=\bigg(-4u^{\dfrac{1}{2}}+\dfrac{u^{\dfrac{3}{2}}}{3}\bigg)\bigg|_0^{\sqrt3}\\\\\\\\\text{Back substitute u}=4-v\\\\\bigg(-4(4-v)^{\dfrac{1}{2}}+\dfrac{(4-v)^{\dfrac{3}{2}}}{3}\bigg)\bigg|_0^{\sqrt3}\\[/tex]
[tex]\text{Back substitute v}=x^2\\\\\bigg(-4(4-x^2)^{\dfrac{1}{2}}+\dfrac{(4-x^2)^{\dfrac{3}{2}}}{3}\bigg)\bigg|_0^{\sqrt3}\\\\\\\text{Solve:}\\\\\bigg(-4(4-3)^{\dfrac{1}{2}}+\dfrac{(4-3)^{\dfrac{3}{2}}}{3}\bigg)-\bigg(-4(4-0)^{\dfrac{1}{2}}+\dfrac{(4-0)^{\dfrac{3}{2}}}{3}\bigg)\\\\\\=\bigg(-4(1)+\dfrac{1}{3}\bigg)-\bigg(-8+\dfrac{8}{3}\bigg)\\\\\\=\bigg(-\dfrac{11}{3}}\bigg)-\bigg(-\dfrac{16}{3}\bigg)\\\\\\=-\dfrac{11}{3}}+\dfrac{16}{3}\\\\\\\large\boxed{\dfrac{5}{3}}[/tex]
