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Calc 2: Integral of 1/sqrt(4-x^2)

  1. Aug 19, 2011 #1
    1. The problem statement, all variables and given/known data
    ∫ from 0 to 1 of 1/sqrt(4-x^2) dx

    I know from Wolfram Alpha that the answer should be π/6, and that the indefinite integral would be arcsin(x/2).
    I see the connection between this and the antiderivative of arcsin, I'm just not sure how to handle the 4 rather than the 1--which technique to use.

    2. Relevant equations
    ∫ 1/sqrt(1-x^2)dx is arcsin x

    3. The attempt at a solution
    Tried a few ways.

    1st attempt was to rewrite as ∫ (4-x^2)^(-1/2)dx then reverse the power and chain rules to get sqrt(4-x^2)/2x from 0 to 1=sqrt(3)/2. Knew this was incorrect but trying to get my thoughts flowing.

    2nd attempt was with integration by parts, with u=1/sqrt(4-x^2) and dv=dx...etc. This one ended up pretty ugly though, as vdu was worse than the original equation.

    3rd attempt was with u-substitution; this one looked like it was going to work then took a bad turn. Let u=4-x^2, produces ∫ from 4 to 3 1/sqrt(u) --> -sqrt(u) from 4 to 3, then plugging back the 4-x^2 ended up undefined.

    I was thinking of partial fractions but the sqrt throws me off. I'm studying for a placement exam and this is a question from a past exam in the class.

    Thanks to everyone very much for your help!
  2. jcsd
  3. Aug 19, 2011 #2


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    Homework Helper

    Write the integrand
    and use t=x/2 as integration variable.

  4. Aug 19, 2011 #3
    ∫ from 0 to 1 of 1/sqrt(4-x^2) dx
    If you have no idea how to evaluate this just use a trigonometric function substitution, if you haven't memorized what this integral is, have you learned about how to make these substitutions yet? It's something you should of learned or well in most calculus II courses.

    This or the option above both work.
  5. Aug 19, 2011 #4


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    ehild's is a good answer. Another way of approaching it is to use the substitution, as GreenPrint suggests, x= 2sin(u). That way [itex]4- x^2= 4- 4sin^2(u)= 4(1- sin^2(u)= 4 cos(u)[/itex].
  6. Aug 19, 2011 #5
    Awesome, thanks guys. Looking at the course's syllabus it looks like they were going for trig substitution here, a topic we didn't cover in my BC class in high school. However, both solutions make sense and now I've learned something new! Thank you again.
  7. Aug 19, 2011 #6
    This is not a good thing. I take it your taking Multivariable Calculus this semester than sense you took the college equivalent of Calculus II in high school? I guess the AP program doesn't directly reflect courses taught in universities. I also noticed that Maclaurin series is taught in AP BC Calculus but not in my college Calculus II course... interesting... the two seem to differ slightly...
  8. Aug 19, 2011 #7
    Exactly that situation. Fortunately my college's Calc II class has the syllabus posted online and it looks like the only discrepancies are trig integrals and substitution and approximate integration. They both cover Maclaurin series, for me. Good luck in your class and thanks again for your help!
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