<s>[INST] 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? 
 [/INST] put dx = 5cost dt
from your fourth step,
5∫cost * 5cost dt = 25∫(cos^2)(t) dt
                  = 25∫(1 + cos2t)/2 dt
                  = (25/2)∫(1 + cos2t)dt
                  = (25/2)(t + (sin2t)/2) + c
                  =(25/2)(t + (2 sint cost)/2) + c

and substitute the value of t and sint and cost in above. Then you will get your answer.
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