Đáp án: $\dfrac13\pi-\dfrac{\sqrt{3}}{2}$
Giải thích các bước giải:
Ta có $4-x^2>0\to x^2<4$
$\to$Đặt $x=2\sin(u)$
$\to dx=d(2\sin u)$
$\to dx=2\cos udu$
Khi đó:
$I=\displaystyle\int \dfrac{x^2}{\sqrt{4-x^2}}dx$
$\to I=\displaystyle\int \dfrac{4\sin^2u}{\sqrt{4-4\sin^2u}}\cdot 2\cos udu$
$\to I=\displaystyle\int \dfrac{4\sin^2u}{\sqrt{4(1-\sin^2u)}}\cdot 2\cos udu$
$\to I=\displaystyle\int \dfrac{4\sin^2u}{\sqrt{4\cos^2u}}\cdot 2\cos udu$
$\to I=\displaystyle\int \dfrac{4\sin^2u}{2\cos u}\cdot 2\cos udu$
$\to I=\displaystyle\int 4\sin^2udu$
$\to I=\displaystyle\int 2\cdot 2\sin^2udu$
$\to I=\displaystyle\int 2\cdot(1-\cos2u)du$
$\to I=2(u-\dfrac12\sin2u)+C$
$\to I=2u-\sin2u+C$
$\to I=2\arcsin(\dfrac12x)-\sin(2\arcsin(\dfrac12x))+C$
$\to \displaystyle\int^1_0 \dfrac{x^2}{\sqrt{4-x^2}}dx=\dfrac13\pi-\dfrac{\sqrt{3}}{2}$
