## anonymous 4 years ago Integration question: $\int\limits_{}^{} {1 \over \sqrt{3x-x^2}}$

1. anonymous

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

2. anonymous

i believe it is linked to arcsin...

3. anonymous

just need a pointer in right direction not full soln

4. ash2326

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

5. anonymous

brilliant

6. anonymous

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

7. anonymous

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

8. anonymous

all right with the rest

9. anonymous

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!