Đáp án:
$I=\dfrac{a^3\pi}{16}.$
Giải thích các bước giải:
$I=\displaystyle\int\limits^a_0 x^2\sqrt{a^2-x^2} \, dx, a>0\\ x=a\sin t \left(t \in \left[-\dfrac{\pi}{2};\dfrac{\pi}{2}\right]\right)\\ dx=a\cos t\\ \begin{array}{|c|c|} \hline x&0&a\\\hline t&0&\dfrac{\pi}{2} \\\hline\end{array}\\ I=\displaystyle\int\limits^\tfrac{\pi}{2}_0 a^2\sin^2t\sqrt{a^2-a^2\sin^2t} \, a\cos t dt\\ =\displaystyle\int\limits^\tfrac{\pi}{2}_0 a^3\sin^2t.\cos^2t \, dt\\ =a^3\displaystyle\int\limits^\tfrac{\pi}{2}_0 \sin^2t\cos^2t\, dt\\ =\dfrac{1}{4}a^3\displaystyle\int\limits^\tfrac{\pi}{2}_0 (2\sin t\cos t)^2\, dt\\ =\dfrac{1}{4}a^3\displaystyle\int\limits^\tfrac{\pi}{2}_0 \sin^2 2t\, dt\\ =\dfrac{1}{4}a^3\displaystyle\int\limits^\tfrac{\pi}{2}_0 \dfrac{1-\cos 4t}{2}\, dt\\ =\dfrac{1}{8}a^3\displaystyle\int\limits^\tfrac{\pi}{2}_0 \left(1-\cos 4t\right)\, dt\\ =\dfrac{1}{8}a^3\left(t-\dfrac{\sin 4t}{4}\right) \Bigg \vert ^\tfrac{\pi}{2}_0 \\ =\dfrac{a^3\pi}{16}.$
