int_(0)^a (x^2sqrt(a^2-x^2)),(rm dx)(a>0)

令$x=a \sin t$，$t\in(0,{\pi\over 2 })$

则$\int_{0}^{a} {x^2\sqrt{a^2-x^2}}\,{\rm dx}$$=\int_{0}^{\pi\over 2} {a^2\sin^2t \cdot a\cos t}\,{\rm dasint}$

$=\int_{0}^{\pi\over 2} {a^3\sin^2t \cdot a\cos^2 t}\,{\rm dt}$

$=\int_{0}^{\pi\over 2} {a^4\sin^2t \cdot \cos^2 t}\,{\rm dt}$

$=\int_{0}^{\pi\over 2} {{a^4\over 4}\sin^22t }\,{\rm dt}$

$={a^4\over 4}\int_{0}^{\pi\over 2} {{1\over 2}(1-\cos 4t)}\,{\rm dt}$

$={a^4\over 8}(t-{1\over 4}\sin 4t)|_0^{\pi\over 2}$

$=$$a^4\pi\over 16$