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Integration by u substitution for inverse trig formulas

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data


    You know the U substitution proofs for inverse trig functions that go like this:

    [itex]\int\frac{1}{a^{2}+x^{2}}dx[/itex]

    [itex]\int\frac{a\frac{1}{a}}{a(1+\frac{x^2}{a^2})}dx[/itex]

    let u = x/a

    du= dx/a

    ....

    [itex]\frac{1}{a}tan^{-1}(x/a)+c[/itex]


    I have searched google and can't find any of these proofs for the other trigonometric functions like sin and tan. Where can I find step by steps of these particular integration generalizations?

    Do these U substitution methods for inverse trig functions have some sort of specific name?
     
  2. jcsd
  3. Sep 28, 2012 #2

    HallsofIvy

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    Last edited by a moderator: May 6, 2017
  4. Sep 29, 2012 #3
    Hi Hallsofivy, really appreciate it. However, of the links you kindly provided, there is only one instance of a formula of which i'm describing. One for arcsin. But there seems to be no resource on the web that displays all or even 2 or 3 of these particular formulas. I've searched every combination of "inverse," "trigonometric," "integration," "subsititution," "formula(s)." I don't know how else to describe these specific formulas. Do you have any other keyword suggestions or links that provide on this integration topic?
     
  5. Sep 29, 2012 #4

    SammyS

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    I have crossed out an extra letter a , which you had in the numerator.
    I don't see how what you have is much of a proof of anything.

    You can get this from taking the derivative of tan-1(x/a).
    [itex]\displaystyle \frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right)=\frac{1}{a}\frac{1}{(x/a)^2+1}[/itex]​

    To do something similar with arcsin & arccos, go ahead and take the derivatives:

    [itex]\displaystyle \frac{d}{dx}\sin^{-1}\left(\frac{x}{a}\right)[/itex]

    [itex]\displaystyle \frac{d}{dx}\cos^{-1}\left(\frac{x}{a}\right)[/itex]
     
  6. Sep 29, 2012 #5

    I didn't mean proof, sorry about that. I guess I just meant formula. Thanks, I'll give it a try.
     
  7. Sep 29, 2012 #6
    [itex]sin^{-1}\frac{x}{a}=y[/itex]

    [itex]\frac{d}{dx}sin^{-1}=\frac{d}{dx}\frac{x}{a}[/itex]

    [itex]cos(y)y'= \frac{1}{a}[/itex]

    [itex]y'=\frac{1}{acos(y)}[/itex]

    [itex]y'=\frac{1}{acos(sin^{-1}(\frac{x}{a})^{2})}[/itex]

    Trig Identity: [itex]sin^{2}y+cos^{2}y=1[/itex]

    cosy=[itex]\sqrt{1-(\frac{x}{a})^2}[/itex]

    [itex]y'=\frac{1}{acos(y)}=\frac{1}{a\sqrt{1-(\frac{x}{a})^{2}}}[/itex]


    So when I do the derivative of arcsine, I get [itex]\frac{d}{dx}sin^{-1}\frac{x}{a}=\frac{1}{a\sqrt{1-(\frac{x}{a})^{2}}}[/itex]

    However, the formula states:

    [itex]\int\frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}\frac{x}{a}+C[/itex]

    I seem to have gotten an "a" outside of the radical and a "1" inside the radical, while the formula shows nothing outside the radical and an "[itex]a^{2}[/itex] inside the radical. Where might I have gone off track to arrive at a final answer with misarranged variables?

    EDIT:

    I think I see it, would it just be a matter of distributing the "a" into the radical?
     
  8. Sep 29, 2012 #7

    SammyS

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    Yes, they are equivalent.
     
  9. Sep 29, 2012 #8
    Thanks SammyS,


    Also,

    You show that the derivative of arctanx is:
    [itex]\displaystyle \frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right)=\frac{1}{a}\frac{1}{(x/a)^2+1}[/itex]​

    But if I am interested in deriving the function [itex]\frac{1}{a}tan^{-1}(\frac{x}{a})[/itex] for this particular formula, does the extra fraction [itex]\frac{1}{a}[/itex] that the function to be derived includes account for the final result of [itex]\frac{1}{a}\frac{1}{a}\frac{1}{(\frac{x}{a})^{2}+1}[/itex] = [itex]\frac{1}{x^{2}+a^{2}}[/itex]?
     
    Last edited: Sep 29, 2012
  10. Sep 29, 2012 #9

    SammyS

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    The algebra says yes.
     
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