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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!

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    Observe you have $\;2u-u^2=-(u-1)^2+1\;$ , *not* what you wrote there...Besides this, your very first integral's limits don't make much sense: you have $\;u\;$ both in the upper limit and **also** it is the integration variable. – 2017-01-05
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    That simple correction helped me solve the integral, thank you! I should probably just an apostrophe on the limit $u$ -> $u'$, but in this case u represents a velocity. – 2017-01-05

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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...