$I=\displaystyle\int\limits_0^{\sqrt{2}} {{\dfrac{x^2}{\sqrt{{4-x^2}}}}}dx$

Hướng dẫn giải:

Đặt $x=2\sin t$

Ta có: $dx=2\cos tdt$

Đổi cận: $x=0\Rightarrow t=0$ và $x=\sqrt{2}\Rightarrow t=\dfrac\pi 4$

 Ta có: $I=\displaystyle\int\limits_0^{\dfrac\pi 4} {{\dfrac{4{\sin }^2t}{\sqrt{{4-4{\sin }^2t}}}2.\cos tdt}}$ $=\displaystyle\int\limits_0^{\dfrac\pi 4} {{\dfrac{8{\sin }^2t.\cos t}{\sqrt{{4{\cos }^2t}}}dt}}$ $=\displaystyle\int\limits_0^{\dfrac\pi 4} {{\dfrac{8{\sin }^2t.cost}{2\cos t}dt}}$

$=\displaystyle\int\limits_0^{\dfrac\pi 4} {{4{\sin }^2tdt}}=2\displaystyle\int\limits_0^{\dfrac\pi 4} {{\left({1-\cos 2t}\right)dt}}$ $=2{\left({t-\dfrac12\sin 2t}\right)}\bigg|_0^{\dfrac\pi 4}$

$=2\left({\dfrac\pi 4-\dfrac12\sin \dfrac\pi 2}\right)-2\left({0-\dfrac12.\sin 0}\right)$ =$\dfrac\pi 2-1.$