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? 

For these types of problems, try to substitute the variable $x$ so that the integral reduces into a form that is easier to integrate.
For this integral, I would start by completing the square in the integrand:
$\sqrt{10x-x^2} = \sqrt{-(x^2-10x)}= \sqrt{-((x-5)^2-25)}=\sqrt{25-(x-5)^2}$.
Note that, so far, I have made no substitution... I have just fiddled around with the integrand.
Now I am ready to substitute. Let $x-5 = t$.
Then the integrand turns into $\sqrt{25-t^2} = 5\sqrt{1-\frac{t^2}{5^2}}$ (which you ought to know how to integrate... either off hand or by further substitution)
and also
$\frac{dx}{dt} = 1 $ which implies $dx = dt$.
