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##\int \frac{1}{\sqrt{1-x^2}} dx##

  1. Dec 27, 2016 #1
    1. The problem statement, all variables and given/known data
    I am doing a little review and having a some trouble deriving the integral ##\int \frac{1}{\sqrt{1-x^2}} dx##

    2. Relevant equations


    3. The attempt at a solution
    Initially I was trying to solve this integral using the substitution ##\cos \theta = x##. I drew my triangle so that the side adjacent to ##\theta## was ##x##, the hypotenuse was ##1##, and from this found that the opposite side was ##\sqrt{1-x^2}##; after this I computed my differentials, performed various substitutions, and concluded that ##\int \frac{1}{\sqrt{1-x^2}} dx = - \arccos x + c##. However, this wrong. So, I tried the substitution ##\sin \theta = x## and I derived the correct formula. Why didn't my first substitution work?
     
  2. jcsd
  3. Dec 27, 2016 #2

    robphy

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  4. Dec 27, 2016 #3

    Ray Vickson

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    Your first expression ##-\arccos(x) + C## is correct; differentiate it to check. That is something that you should always do.

    Think about why there are two different-looking answers!
     
  5. Dec 27, 2016 #4

    vela

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    If you can't figure out what Ray's getting at, try this: set C=0 for simplicity and plot the two results.
     
  6. Dec 27, 2016 #5

    robphy

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    Call the integral ##\theta\equiv \int \frac{1}{\sqrt{1-x^2}} dx##.
    Then we have
    ##\theta=-\arccos(x) + C_1## and ##\theta=\arcsin(x) + C_2##,

    Something that is sometimes forgotten is that the constants of integration (##C_1## and ##C_2## ) are not generally equal.
    That is, if we choose a value for ##C_1##, what is the corresponding choice for ##C_2##?
    Let ##C_2=C_1+Q##, where ##Q## is an unknown.

    In this case, it might help interpreting the results as follows:
    ##\cos(C_1-\theta)=x ## and ##\sin(\theta-C_2)=x##.
    What is ##Q##?
     
  7. Dec 27, 2016 #6

    robphy

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