## Thomas' Calculus 13th Edition

$sin^{-1}(\theta-1)+C$
$\int \frac{d\theta}{\sqrt {2\theta-\theta^{2}}}=\int\frac{d\theta}{\sqrt {1-(\theta-1)^{2}}}$ Put $\theta-1=x$ Then $\frac{dx}{d\theta}= 1$ or $dx=d\theta$ Therefore, $\int \frac{d\theta}{\sqrt {2\theta-\theta^{2}}}=\int \frac{dx}{\sqrt {1-x^{2}}}=sin^{-1}(\frac{x}{1})+C$ $=sin^{-1}(\theta-1)+C$