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? 
 A: 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.
A: 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.
A: 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 $dx = dt$.
A: As Thomas said $\displaystyle \frac{dx}{dt} = 1$ not $5dt$.
If $\displaystyle x - 5 = t$ then $\displaystyle x = t - 5 \therefore \frac{dx}{dt} = 1$.
But what you really should do, is use the sustitution originally suggested and remeber to integrate with respect to $t$.
A: 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.
