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Find the following indefinite integral

  1. Apr 1, 2012 #1
    1. The problem statement, all variables and given/known data

    find the indefinite integral of ∫x√(1-x^2) dx

    2. Relevant equations



    3. The attempt at a solution
    ∫x√(1-x^2) dx

    let x = sinθ
    dx = cosθ dθ

    now sin^2θ + cos^2θ = 1
    => cosθ = √1-sin^2θ ( for the form √(1-x^2))

    ∫x√(1-x^2) dx => ∫sinθ (√1-sin^2θ) cosθ dθ
    => ∫sinθ cos^2θ dθ

    integrating :- let u = cosθ
    du = -sinθ dθ

    => ∫ sinθ cos^2θ dθ = ∫ -u^2 du
    => -1/3 u^3 + c
    => -1/3 cos θ + c


    is this correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 1, 2012 #2

    SammyS

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    Hello jodecy. Welcome to PF !

    At the very least, you need to give the final result in terms of the variable, x .

    A more direct substitution is to let u = 1-x2 .
     
  4. Apr 1, 2012 #3
    ohhhh so i need to state θ = arcsinx

    therefore my answer in terms of x would be -1/3 cos^3(arcsinx )?

    when i come back i'll try subst with u = 1-x^2
     
  5. Apr 1, 2012 #4
    Well, I wouldn't even mess with trig substitution, as SammyS said; that's just overkill and too much work for this problem.

    But, it is also not "correct" in the sense that it is really hard to understand what [itex](-1/3)\cos^3(\arccos(x))[/itex]. So, here is a better way.

    Draw a right triangle. Pick one of the non-right angles to be [itex]\theta[/itex]. Now, you now that [itex]\sin (\theta)=x[/itex], and you know that sine is opposite over hypotenuse. Now, label the opposite side with an [itex]x[/itex] and the hypotenuse with a [itex]1[/itex]. Now, do you see how to find the cosine of [itex]\theta[/itex] out of that?
     
  6. Apr 2, 2012 #5
    ok so i worked it out using the subst of u = 1 - x^2 and got the answer to be
    -1/3 (1-x^2) ^ 3/2 + c. however the original method i used was that presented to me by my lecturer so i'm wondering ...
     
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