<s>[INST] How are these two integrals related? How to express the integral $$\int_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx $$ 
in terms of the integral $$ \int_{-1}^{1} \sqrt{1-x^2} \ dx?$$ I know that the latter integral is equal to $\pi / 2$. 
We can't use substitution. We can only use the following two results: 


*

*$$ \int_{a}^{b} f(x) \ dx = \int_{a+c}^{b+c} f(x-c) \ dx.$$ 


And


*

*$$ \int_{a}^{b} f(x) \ dx = \frac{1}{k} \int_{ka}^{kb} f(\frac{x}{k}) \ dx$$ for any $k \neq 0$. 

 [/INST] Using $k=\frac{1}{2}$ in the second rule, we get
$$\begin{align}
\int\limits_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx &=2\int\limits_{-1}^{1} (2x-3) \sqrt{4-4x^2} \ dx
\\&=2\int\limits_{-1}^{1} 2(2x-3) \sqrt{1-x^2} \ dx
\\&=4\int\limits_{-1}^{1} (2x-3) \sqrt{1-x^2} \ dx
\\&=8\int\limits_{-1}^{1}x\sqrt{1-x^2} \ dx-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ dx
\\&=-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ dx
\end{align}
$$
where the last equality occurs because $x\sqrt{1-x^2}$ is an odd function and we are integrating from $-1$ to $1$, so $\int_{-1}^{1}x\sqrt{1-x^2} \ dx=0$.
</s>