Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do you integrate this?

  1. Feb 11, 2007 #1
    How do you integrate this??

    Integral of 3x^2 + (4-x^2)^(1/2) dx ??

    I tried a u substitution for the 4-x^2 but what do you do with 3x^2? If someone could walk we through this, i'd greatly appreciate it...i hate being stuck on something so trivial in the middle of a problem.
  2. jcsd
  3. Feb 11, 2007 #2
    Break it into two integrals, 3x^2 and (4-x^2)^(1/2).

    Then use maybe x=2sin(u)
  4. Feb 11, 2007 #3
    sin??? where did sin come into the picture? Is that the only way to do this???
  5. Feb 11, 2007 #4
    [tex]\int 3x^2 + \sqrt{4-x^2} \,dx[/tex]

    Yes, break into two integrals:

    [tex]\int 3x^2 \, dx + \int (4-x^2)^\frac{1}{2} \, dx[/tex]

    I think theperthvan is saying the second integral needs trig substitution.
    Last edited: Feb 11, 2007
  6. Feb 11, 2007 #5
    so is the second part equal to (2/3)(4-x^2)^3/2 ?? you don't have to do anything with the internal part, 4-x^2 like you do with the product rule in differentiation?
  7. Feb 11, 2007 #6
  8. Feb 11, 2007 #7
    No, that will most definitely not work. Here you have a polynomial under a square root. You need to get rid of the square root. Here's what you do.

    For [tex]\int \sqrt{a^{2}-x^{2}} \,dx[/tex], use [tex]x = a \, sin(\theta)[/tex]

    Note that you also need to substitute the differential, [tex]dx = a \, cos(\theta)d\theta[/tex]

    I hope I'm not saying too much, but also remember to use a certain trigonometric identity to eliminate the radical.

    Having said this, if you don't know what a trigonometric substitution is, then my guess is that your calculus class hasn't yet covered it. Do you need to compute an antiderivative, or are you trying to compute a definite integral? Because if you're doing the definite integral from 0 to 2, you can do this simply by using the formula for the area of a circle.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook