<s>[INST] How do I calculate $\int_0^ax^2\sqrt{a^2-x^2}dx$ via substitution? I have to calculate the integral $$\int_0^ax^2\sqrt{a^2-x^2}dx$$ using solely substitution (no integration by parts). $a$ is a positive constant.
I'm confused on how to do this?
 [/INST] Let $x=a\sin\theta\implies dx=a\cos \theta\ d\theta$
$$\int_0^a x^2\sqrt{a^2-x^2}\ dx=\int_{0}^{\pi/2} a^2\sin^2\theta(a\cos\theta)(a\cos \theta\ d\theta)$$
$$=a^4\int_{0}^{\pi/2}  \sin^2\theta\cos^2\theta\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2}  (2\sin\theta\cos\theta)^2\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2} \sin^22\theta\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2} \frac{1-\cos4\theta}{2}\ d\theta$$
$$=\frac{a^4}{8}\int_{0}^{\pi/2} (1-\cos4\theta)\ d\theta$$
$$=\frac{a^4}{8}\left(\frac{\pi}{2}-0\right)=\frac{\pi a^4}{16}$$
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