# 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:

$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\ = \int \sqrt{50+50\sin{t} - (25+50\sin{t} + 25\sin^2{t})} dx\\ = \int \sqrt{ 25-25\sin^2{t} } dx \\ = 5 \int \sqrt{1-\sin^2{t}} dx \\ = 5\sin^{-1}{\sin{t}} \\ = 5t \\ = 5 \sin^{-1}{\frac{x-5}{5}}$

$$\frac{25}{2}\sin^{-1}{\frac{x-5}{5}}+\frac{x-5}{2}\sqrt{10x-x^2}+c$$

What did I do wrong? Or is the answer wrong perhaps?

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I don't think that was so much a "hint" as an "Apply x as $5+5\sin{t}$". – Neil Apr 24 '12 at 9:44
If you substitute $x=5+5sin(t)$, then you must note that $dx=5cos(t)dt$, which you forgot from your calculations. Right now you're integrating $t$ respect to the variable $x$. – Thomas E. Apr 24 '12 at 9:46

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.

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 Hmm ... so am I right to say: $$5 \int \sqrt{1-\sin^{-1}{t}} \cdot \cos{t} dt$$ Then I do integration by parts, Let $u = \cos{t} \Rightarrow du = -\sin{t} dt$ Let $dv = \sqrt{1-\sin^2{t}} dt \Rightarrow v=t$ $$= 5[t\cos{t} + \int t\sin{t} dt]$$ Then integration by parts again, this time ... Let $u=t \Rightarrow du = dt$ Let $dv = \sin{t} dt \Rightarrow v = -\cos{t}$ $$=-t\cos{t}+\int \cos{t} dt = -t\cos{t} + \sin{t}$$ $ori eq = 5\sin{t} +c$? Yet another answer? – Jiew Meng Apr 24 '12 at 10:50 Let's start from $\int \sqrt{25-y^2}\, dy$, as I told you. Put $y=5\sin t$, $dy=5 \cos t \, dt$. Now $$\int \sqrt{25-y^2}\, dy = \int \sqrt{25-25\sin^2 t} \cdot 5 \cos t \, dt = 25 \int \cos^2t \, dt.$$ Try to complete. – Siminore Apr 24 '12 at 11:04

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.

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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 $dx = dt$.

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 Ok, So I have ... $$5\int\sqrt{1-(\frac{t}{5})^2}=5\sin^{-1}{\frac{x-5}{5}}$$ Is it right? I also tried letting $t=\frac{x-5}{5}, dx=5 dt$ $$5 \int \sqrt{1-t^2}\cdot 5 dt=25 \sin^{-1}{\frac{x-5}{5}}$$ They appear different? I must me making some very stupid mistakes again?  – Jiew Meng Apr 24 '12 at 10:27 @Jiew Check your integral tables. You're confusing the integral you want to solve with $$\int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1} x+C$$ – Peter Tamaroff Apr 24 '12 at 12:42

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 $t$.

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 I get $dx = 5dt$ when I let $t=\frac{x-5}{5}$. Thats 1 possible substitution too? – Jiew Meng Apr 24 '12 at 11:07 Yes you're right, I didn't see that, but I'm not sure that's the way to go. – Billy Ray Valentine Apr 24 '12 at 11:13

put dx = 5cost dt

5∫cost * 5cost dt = 25∫(cos^2)(t) dt