How to $\int \sqrt{10x-x^2}dx$ Theres a hint to use $x=5+5\sin{t}$. Ok, but how do I know what substitution to use if a hint wasn't given? Is it "trivial" or perhaps, its very unlikely that that will appear? 
Anyways, I did: 
$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\
= \int \sqrt{50+50\sin{t} - (25+50\sin{t} + 25\sin^2{t})} dx\\
= \int \sqrt{ 25-25\sin^2{t} } dx \\
= 5 \int \sqrt{1-\sin^2{t}} dx \\
= 5\sin^{-1}{\sin{t}} \\
= 5t \\
= 5 \sin^{-1}{\frac{x-5}{5}}$
But the answer was: 
$$\frac{25}{2}\sin^{-1}{\frac{x-5}{5}}+\frac{x-5}{2}\sqrt{10x-x^2}+c$$
What did I do wrong? Or is the answer wrong perhaps? 

What you "did wrong" in making the subsitution was forget to take into account the
$$\frac{dx}{dt} = 5\cos t.$$ 
Continuing,
$$\begin{align}
\int \sqrt{10x-x^2}dx & = \int \sqrt{10(5 + 5\sin{t}) + (5+5\sin{t})^2}5\cos{t}dt\\  
& = 5\int \sqrt{25 - 25\sin^2t}\cos{t}dt\\  
& = 25\int \cos^2tdt\\
& = \frac{25}{2}\sin{t}\cos{t} + \frac{25}{2}t\\
& = \frac{25}{2}\frac{(x-5)}{5}\sqrt{1-(\frac{x-5}{5})^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5})\\
& = \frac{x-5}{2}\sqrt{25 - (x-5)^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5})\\
& = \frac{x-5}{2}\sqrt{10x-x^2} + \frac{25}{2}\sin^{-1}(\frac{x-5}{5}).
\end{align}$$
Not the easiest method, but it does work.
