How do we obtain $\int_{0}^{1}x\sqrt{1+4x^2}\,dx$ from $\displaystyle\int_{1}^{2}\sqrt{(2-y)+\frac{1}{4}}\,dy$ using $u$ substituition?
My approach :
Let $u^2=2-y$. Then $2u\,du=-dy$ and when $y=1$, $u=1$, and when $y=2$, $u=0$.
So \begin{align} \int_{1}^{2}\sqrt{(2-y)+\frac{1}{4}}\,dy&=\int_{1}^{0}\sqrt{u^2+\frac{1}{4}}(-2u)\,du\\ &=\int_{0}^{1}2u\sqrt{u^2+\frac{1}{4}}\, du\\ &=\int_{0}^{1}u\sqrt{4u^2+1}\,du\\ &=\int_{0}^{1}x\sqrt{4x^2+1}\,dx. \end{align}
But is this kind of $u$-substitution possible?
I thought we usually substitute $u=$ something not $u^2=$ something.