I'm trying to solve the following definit integral.
$$ \int_0^u \dfrac{\mathrm{d}u}{\sqrt{2u-u^2}} $$ My procedure thus far has been to first complete the square then make a change of variable $v=u-1$ and $ \cos(w)=v$. $$ \int_0^u \dfrac{\mathrm{d}u}{\sqrt{(u-1)^2-1}} = \int_{-1}^{v+1} \dfrac{\mathrm{d}v}{\sqrt{v^2-1}} = \int_{\pi}^{\cos(w)} \dfrac{-\sin(w)\mathrm{d}w}{\sqrt{cos^2(w)-1}} = \int_{\pi}^{\cos(w)}i\mathrm{d} w$$
Since the upper limit in the first integral is $u$, I'm not sure what the new upper limit will be when I change variable. I'm also unsure if this is the correct way to solve the integral that's why I would appreciate help from this community!
Many thanks!
My idea, which I think is much simpler:
$$\int\frac{du}{\sqrt{2u-u^2}}=\int\frac{du}{\sqrt{1-(u-1)^2}}=\arcsin(u-1)+K\text{ (=constant)}$$
Without change of variable or anything: the above is an immediate integral. You just need to check carefully your integral's limits...