<s>[INST] Evaluating $\int^1_0x\ \sqrt{1-x^2}dx$ How do I find the following integral:
$$\int^1_0x\ \sqrt{1-x^2}dx$$
 [/INST] $$\int^1_0x\ \sqrt{1-x^2} = \int^1_0 \ x(1 - x^2)^{1/2}dx$$
Let $u = 1 - x^2.\;$ Then $du = -2x\,dx \iff x\,dx = -\frac 12\,du$
Bounds of integration: At $x = 0, u = 1,\;$ and at $x = 1, u = 0$
Substituting gives us: $$
\begin{align} \int^1_0\ x\ (1-x^2)^{1/2}\,dx 
& =  \int_1^0\ u^{1/2}\left(-\frac 12\,du\right) \\ \\
& = -\frac 12 \int_1^0 \ u^{1/2}\,du \\ \\
& = -\frac 12 u^{3/2}\Big|_1^0 \\ \\
& = \frac 12 u^{3/2}\Big|_0^1 \\ \\
& = \frac 12(1 - 0) = \frac 12
\end{align}
$$
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