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Arcsin Integral Format

  1. May 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral ∫1/(x^4)(sqrt(a^(2) - x^(2))


    2. Relevant equations
    u= dx/sqrt(a^2-x^2)
    du= arcsin(x/a)dx

    3. The attempt at a solution

    Am I even doing this right? And if so, how would I express 1/x^4 in terms of u? Have I taken the right u-substitution? Thanks.
     
  2. jcsd
  3. May 26, 2013 #2

    Simon Bridge

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    Lets make sure I'm reading that right:
    You are asked to evaluate:
    $$\int \frac{dx}{x^4\sqrt{a^2-x^2}}$$ ... and you are using the substitution:
    $$u=\frac{dx}{\sqrt{a^2-x^2}}\\ du = \arcsin(x/a)dx$$ ... I have no idea why you would choose such a substitution or how you arrived at it.

    The usual way to get rid of a square root would be to use a trig substitution like ##x=a\sin\theta## so that ##dx = a\cos\theta\; d\theta##
     
  4. May 26, 2013 #3
    I chose this substitution because one of the formulas I learnt is that the integral of dx/sqrt(a^2-x^2) is arcsin(x/a) dx; but it's supposed to be the derivative. My mistake :/

    Using trig substitution, I get:

    int [1/(x^4)(sqrt(a^2-sin^2(theta))]

    I'm not sure how to continue at all.
     
  5. May 26, 2013 #4

    Simon Bridge

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    int [1/(x^4)(sqrt(a^2-sin^2(theta))] would be if you put ##x=\sin\theta## ... that's the wrong substitution.

    You cant do just any old substitution.

    Here, what you want to do is exploit that ##\sin^2\theta + \cos^2\theta = 1##.

    Don't forget that the substitution applies to every instance of x and to dx also.
     
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