Sqrt Integration

Homework Statement

hi guys ,, how are you all ,,
i got another problem -_-

integral(1/sqrt(4 x-x^2), x)

The Attempt at a Solution

i have no idea , i tried and lift it up and make it (4x-x^2)^-0.5 and i even took x as common factor x^-0.5 * (4-x)^-0.5 but still couldn't do anything next ,, any ideas ??

quantumdude
Staff Emeritus
Gold Member
You need to complete the square under the radical sign.

You need to complete the square under the radical sign.

i don't think they taught us how to change it to radicals ,, can you tell me what's the name of the chapter that teach this thing ? or at least give me the name of the method so i can look up for it

djeitnstine
Gold Member
Lord dark, complete the square. That will get you to the next step

Edit: tom beat me to it

edit2: dark, he did not say change it radicals, but to complete the square

lol ,, got the idea ,, thanks guys ,, i'll try then i'll give you the results

i tried to solve and i reached here ,, 1/sqrt(4x-x^2) = 1/abs(x)*sqrt(4-x^2) and i know it has something with inv(sec x) but the problem now is 4 ,, i don't know how to make an equation that differentiable and give 1/sqrt(4x-x^2) from inv (sec x),, so i read in wikipedia (http://upload.wikimedia.org/math/6/3/f/63ff35f597f9092362ae154641cfe48a.png)
so i think i get this answer: 0.5*inv(sec (x/2))+c ,, is it right ??

djeitnstine
Gold Member
No that is wrong

When completing the square, you make it in the form $$\left( x \pm a\left)^2 - b$$

So you will be looking at $$\int{\frac{dx}{\sqrt{\left( x \pm a\left)^2 - b}}}$$ which becomes an easy trig substitution

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ok ,, i think i did it right this time : integral[1/sqrt(4-(x-2)^2)] and i know its like the inverse of sin(x) [http://upload.wikimedia.org/math/2/e/8/2e805e2888bd24e7f61e54a002462204.png] but , i have 4 in square root instead of 1 what should i do ?? i tried and applied the previous way i learn from Wikipedia [0.5*inv(sin((x-2)/2))] but when i check i get wrong answer

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HallsofIvy
$$\frac{1}{\sqrt{4- (x-2)^2}}= \frac{1}{\sqrt{4(1-\frac{(x-2)^2}{4}}}= \frac{1}{2}\frac{1}{\sqrt{1- \frac{(x-2)^2}{4}}}$$
$$\frac{1}{\sqrt{4- (x-2)^2}}= \frac{1}{\sqrt{4(1-\frac{(x-2)^2}{4}}}= \frac{1}{2}\frac{1}{\sqrt{1- \frac{(x-2)^2}{4}}}$$