Answer
$$\sin^{-1}\left(\frac{x-6}{6}\right)+C$$
Work Step by Step
Given $$\int \frac{d x}{\sqrt{12 x-x^{2}}}$$ Since $$\int \frac{d x}{\sqrt{12 x-x^{2}}}= \int \frac{d x}{\sqrt{36-(x-6)^{2}}}$$ Let $$ x-6=6\sin u\ \ \ \ \ \ \ dx=6\cos udu$$ Then \begin{align*} \int \frac{d x}{\sqrt{12 x-x^{2}}}&=\int \frac{d x}{\sqrt{36-(x-6)^{2}}}\\ &= \int \frac{6\cos udu}{\sqrt{36-36\sin^{2}u}}\\ &=\int \frac{6\cos udu}{\sqrt{36(1-\sin^{2}u)}}\\ &=\int du\\ &=u+C\\ &=\sin^{-1}\left(\frac{x-6}{6}\right)+C \end{align*}
