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I came across this $$\int_0^6 \frac{1}{\sqrt{x(6-x)} }dx$$

I thought of completing the square and coming somehow to $$ \frac{1}{\sqrt{1+x²}} $$ which has an antiderivative

Unfortunately this did not work. I guess, this is an indefinite integral, because f(6) is not defined. So personally I think, I have to use lim for this. But for that, I have to have the antiderivative. Does anyone have a tip for me to solve this ? Thank you in advance!

Please look at the edit. The root was missing, I edited it. Please refresh your answers. Thank you.

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    And finding the zero of the function, right? – 2017-01-13
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    Should the denominator in the integral be under a square root sign? The title seems to indicate it, as does your thought of $\frac 1{\sqrt{1+x^2}}$ – 2017-01-13
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    This integral is a divergent one ! – 2017-01-13
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    Sorry, I edited the question! – 2017-01-13
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    As a quick note: the term here is 'improper' integral, not 'indefinite' integral. Indefinite integral generally is used as a synonym for finding the class of antiderivatives for a function. – 2017-01-13

6 Answers 6

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You can write $\int_0^6 \frac{1}{\sqrt{x(6-x)} }dx=\int_0^6 \frac{1}{\sqrt{9-(x-3)^2} }dx$, then $u=x-3$ which looks a lot like what you were trying.

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new version of the question

I'll start with a hint: make a change of variables $$x = 6 \sin^2 u$$ for $u\in [0,\pi/2]$

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    I am sorry, but I forgot the sqrt in f(x). I made an edit. – 2017-01-13
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use that $$\frac{1}{x(6-x)}=-1/6\, \left( -6+x \right) ^{-1}+1/6\,{x}^{-1}$$ for the new integral use this here https://en.wikipedia.org/wiki/Euler_substitution

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    I am sorry, but I forgot the sqrt in f(x). I made an edit. – 2017-01-13
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As the hint by Joe Johnson says, you can use partial fractions to rewrite it like so: $$\dfrac{1}{x(6-x)}=\dfrac{A}{x} + \dfrac{B}{6-x}\implies 1=A(6-x)+Bx$$ choose $x=0$ and $x=6$ to solve for $A$ and $B$ and you should get two integrals that use the natural log. On the other hand, to do your other integral, $\frac{1}{\sqrt{1+x^2}}$, use $x=\tan(u)\implies dx=\sec^2(u)du$. So, $$\int\dfrac{1}{\sqrt{1+x^2}}dx=\int\dfrac{1}{\sqrt{1+\tan^2(u)}}\sec^2(u)du=\int\sec(u)du$$ which is a popular integral that you can do by parts or by partial fractions if you're clever.

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You can write $$\frac{1}{x(6-x)}$$ as $$\frac{1}{x(6-x)} = \frac{A}{x} + \frac{B}{(6-x)}$$

Then, if you equalize denominators, $$\frac{A(6-x)}{x(6-x)}+ \frac{Bx}{x(6-x)}= \frac{1}{x(6-x)}$$

You can proceed to this, $$6A-6x+Bx = 1$$ Where $$A = \frac{1}{6}$$ $$B = 1$$ Which will lead you to $$\frac{1}{x(6-x)}= \frac{1}{6x} + \frac{1}{(6-x)}$$ Now it is easier to find its anti-derivative. Hint: ln function

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We first expand:

$$\int\frac{1}{\sqrt{x(6-x)}}~dx=\int\frac{1}{\sqrt{6x-x^2}}~dx$$

We complete the square:

$$\int\frac{1}{\sqrt{9-(x-3)^2}}~dx$$

We substitute $u=x-3$ and $du=1~dx$

$$\int\frac{1}{\sqrt{9-u^2}}~du=\int\frac{1}{3\sqrt{1-\frac{u^2}{9}}}~du$$

We can now substitute $s=\frac{u}{3}$ and $ds=\frac{1}{3}~du$

$$\int\frac{1}{\sqrt{1-s^2}}~ds=\arcsin(s)+C$$

Resubstituting the variables:

$$\int\frac{1}{\sqrt{x(6-x)}}~dx=\arcsin\left(\frac{x-3}{3}\right)+C$$

Now just substitute your bounds between $0$ and $6$.