anonymous
  • anonymous
Integration question: \[\int\limits_{}^{} {1 \over \sqrt{3x-x^2}}\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i've tried allsort of techniques not sure how to go about
anonymous
  • anonymous
i believe it is linked to arcsin...
anonymous
  • anonymous
just need a pointer in right direction not full soln

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ash2326
  • 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
anonymous
  • anonymous
brilliant
anonymous
  • anonymous
First of all complete the square in the denominator as : \[\int\limits 1/\sqrt{3x - x^2} dx = \int\limits 1/\sqrt{ [9/4-(x-3/2)^2]} \]
anonymous
  • anonymous
Now, For the integrand 1/sqrt(9/4-(x-3/2)^2), substitute u = x-3/2 and du = dx
anonymous
  • anonymous
all right with the rest
anonymous
  • 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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.