Question

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

Answer 1

i've tried allsort of techniques not sure how to go about

Answer 2

i believe it is linked to arcsin...

Answer 3

just need a pointer in right direction not full soln

Answer 4

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

Answer 5

brilliant

Answer 6

First of all complete the square in the denominator as :
\[\int\limits 1/\sqrt{3x - x^2} dx = \int\limits 1/\sqrt{ [9/4-(x-3/2)^2]} \]

Answer 7

Now, For the integrand 1/sqrt(9/4-(x-3/2)^2), substitute u = x-3/2 and du = dx

Answer 8

all right with the rest

Answer 9

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!