<s>[INST] Help with trig sub integral I've gone over my work twice, and while I dropped a negative the first time, I believe that my work is sound. Webwork is not accepting it, however.
Here is the problem and my solution. Sorry for bad handwriting

 [/INST] Let $x=2\sin{t}$.
Thus, $$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{x^2-4+4}{\sqrt{4-x^2}}dx=4\int\frac{1}{\sqrt{4-x^2}}dx-\int\sqrt{4-x^2}dx=$$
$$=4\arcsin\frac{x}{2}-4\int\cos^2tdt=4\arcsin\frac{x}{2}-2\int(1+\cos2t)dt.$$
Can you end it now?
Also, let $x-2=v$.
Thus, we can use the similar way: $$\int\frac{x^2}{\sqrt{4x-x^2}}dx=\int\frac{x^2}{\sqrt{4-(x-2)^2}}dx=\int\frac{v^2+4v+4}{\sqrt{4-v^2}}dv=$$
$$=\int\frac{v^2-4+4v+8}{\sqrt{4-v^2}}dv=\int\frac{4v}{\sqrt{4-v^2}}dv-\int\sqrt{4-v^2}dv+8\int\frac{1}{\sqrt{4-v^2}}dv.$$
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