Integration question: \[\int\limits_{}^{} {1 \over \sqrt{3x-x^2}}\]

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Integration question: \[\int\limits_{}^{} {1 \over \sqrt{3x-x^2}}\]

Mathematics
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i've tried allsort of techniques not sure how to go about
i believe it is linked to arcsin...
just need a pointer in right direction not full soln

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Other answers:

write 3x-x^2 as \(0-(x^2-3x)\) now complete the square \[(0-(x^2-3x+9/4-9/4)\] \[{(3/2)}^2-{(x-3/2)}^2\] now substitute x-3/2 as u dx = du now use the standard formula
brilliant
First of all complete the square in the denominator as : \[\int\limits 1/\sqrt{3x - x^2} dx = \int\limits 1/\sqrt{ [9/4-(x-3/2)^2]} \]
Now, For the integrand 1/sqrt(9/4-(x-3/2)^2), substitute u = x-3/2 and du = dx
all right with the rest
The integral now becomes 1/sqrt(9/4-u^2) du. The integral of 1/sqrt(9/4-u^2) is sin^(-1)((2 u)/3) Now you can proceed easily!

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