Sqrt Integration

[Lord Dark]

Homework Statement

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

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

Homework Equations

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]

You need to complete the square under the radical sign.

 

[Lord Dark]

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]

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

 

[Lord Dark]

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

 

[Lord Dark]

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]

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

 

[Lord Dark]

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

 

[HallsofIvy]

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

 

[Lord Dark]

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

Lol ,, i think am stupid now -_- ,, i got it until the second phase but i didn't think of getting 4 out of square root ,, thanks very much guys for the help