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? 

Try to complete the square: $10x-x^2=10x-x^2+25-25 = -(x-5)^2+25$. After a change of variables $y=x-5$, your integral becomes $\int \sqrt{25-y^2}\, dy$. And yhis should be familiar to you.
