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Trigonometric substitution - Why?

  1. Sep 25, 2010 #1
    Trigonometric substitution - Why???

    Hey guys

    Im sitting here with trigonometric substitution problems, and I have a kind of a problem.

    I can't see WHY it is legal to substitute x for a sin ([tex]\theta[/tex])

    If you have a the integral:

    [tex]\int[/tex][tex]\frac{1}{\sqrt{1-x^2}}[/tex]dx

    Then I know the substitution would be

    x = sin([tex]\theta[/tex])

    But from where I see it, and I guess that is the problem, if the integral is just provided in the form it is above, then x [tex]\in[/tex][tex]\Re[/tex]

    Then why can we substitute x by the function g([tex]\theta[/tex]) = sin ([tex]\theta[/tex]) when
    g([tex]\theta[/tex]) [tex]\in[/tex] [-1 ; 1]. From my point of view there is a problem when the function we substitute by only gives and small fraction of the possible inputs x can have...

    Anyone follow my thoughts? Someone can explain it?

    regards,
    Rune
    Engineering Student
     
  2. jcsd
  3. Sep 25, 2010 #2
    Re: Trigonometric substitution - Why???

    Well to me this technique never seemed anything more than a trick either.

    However, it makes a hell of a lot of sense if you think of a triangle:

    [PLAIN]http://img401.imageshack.us/img401/3499/triangleb.jpg [Broken]

    [tex] \int_{a}^{b}\frac{1}{\sqrt{1 \ - \ x^2}} \,dx [/tex]

    [tex] \int_{a}^{b} \ csc( \theta ) \ dx[/tex]

    Now, looking at the picture we see that:

    [tex] cos( \theta ) \ = \ \frac{x}{1} [/tex]

    [tex] cos( \theta ) \ = \ x[/tex]

    [tex] sin( \theta ) d \theta \ = \ dx [/tex]

    [tex] \int_{a}^{b} \ csc( \theta ) \ dx[/tex]

    [tex] \int_{a}^{b} \ csc( \theta ) \ sin( \theta ) d \theta[/tex]

    [tex] \int_{a}^{b} \ \frac{1}{sin( \theta )} \cdot \ sin( \theta ) d \theta[/tex]

    [tex] \int_{a}^{b} \ \frac{sin( \theta )}{sin( \theta )} d \theta[/tex]

    [tex] \int_{a}^{b} \ 1 d \theta[/tex]

    [tex] \int_{a}^{b} \ d \theta[/tex]

    Remember that when you take the integral you have to get it out of trigonometric form
    and into the standard x's etc... you started with. It will involve arctangents or whatever
    comes up, you know what I mean...

    By the way, this method is extremely helpful in most "fraction-like" integrals, you can
    play with your function, i.e. using completing the square under a radical sign or something,
    to get really complex looking things to be very simple. Enjoy!
     
    Last edited by a moderator: May 4, 2017
  4. Sep 25, 2010 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Trigonometric substitution - Why???

    You main objection "From my point of view there is a problem when the function we substitute by only gives and small fraction of the possible inputs x can have..." simply isn't correct. It is NOT the case that x can be any real number.


    In that particular problem,
    [tex]\int \frac{1}{\sqrt{1- x^2}}dx[/tex]
    x can't be "any real number". Since [itex]1- x^2[/itex] is inside a square root, that is possible only if [itex]1- x^2\ge 0[/itex] which is the same as saying that [itex]x^2< 1[/itex] and that immediately leads to [itex]-1\le x\le 1[/itex].

    We use such trig substitutions to get rid of square roots- and those square roots force the x values to be appropriate for the trig substitution.

    For example, if an integrand involved [itex]\sqrt{a^2- x^2}[/itex], I would use the substitution [itex]x= a sin(\theta)[/itex] so that [itex]\sqrt{a^2- x^2}= \sqrt{a^2- a^2sin^2(\theta)}[/itex][itex]= |a|\sqrt{1- sin^2(\theta)}= |a|cos(\theta)[/itex].

    Since [itex]sin(\theta)[/itex] must be between -1 and 1, [itex]x= a sin(\theta)[/itex] must be between -a and a: exactly what is required for [itex]a^2- x^2\ge 0[/itex] so that [itex]\sqrt{a^2- x^2}[/itex] exists!
     
    Last edited by a moderator: Sep 25, 2010
  5. Sep 25, 2010 #4
    Re: Trigonometric substitution - Why???

    Thank you sponsoredwalk. That was a good explanation! :)

    And HallsofIvy, that is my mistake. Im completely gone, hehe. Didn't see that one. Thanks. Now I get it :)
     
  6. Oct 5, 2010 #5
    Re: Trigonometric substitution - Why???

    I became very interested in sponsoredwalk's method of doing the substitution.
    It is more direct than using x = sin theta to simplify 1 - x^2, and appears quite versatile.
    BUT
    I found that for some integrals the result is wrong.
    Then I noticed that EVEN sponsoredwalk's example appears incorrect to me !!

    Look and see that the integral works out to be "theta"
    The triangle shows cos theta = x so arccos x = theta.
    But the correct antiderivative is arcsin x NOT arccos x.

    Can someone explain why the method fails and the result incorrect ?

    Note that if one changes the triangle so sin theta = x
    it comes out right.

    Thanks for any inputs or comments
     
    Last edited by a moderator: May 5, 2017
  7. Oct 5, 2010 #6
    Re: Trigonometric substitution - Why???

    I'm sorry, if you look at my original derivation I set x = cosθ. When I took the derivative of
    cosθ I forgot to include the minus sign when writing in LaTeX. If you include the minus sign
    then you get the correct answer.

    [tex] - \ \int \ d \theta \ = \ - \ \theta \ + \ C [/tex]

    If you look at the picture we can see that: [tex] \cos \ \theta \ = \ x [/tex]


    [tex] \ \theta \ = \ \cos^{-1} \ ( \ x \ ) \ = \ arccos \ ( \ x \ )[/tex]

    So plugging this all in we get:

    [tex] - \ \int \ d \theta \ = \ - \ \theta \ = \ - \ arccos \ ( \ x \ ) \ + \ C[/tex]

    Still not convinced?

    [tex] y \ = \ - \ arccos \ ( \ x \ ) [/tex]

    [tex] cos \ y \ = \ - \ x \ [/tex]

    [tex] - \ \sin \ y \ \frac{dy}{dx} \ = \ - \ 1 [/tex]

    [tex] - \ \frac{dy}{dx} \ = \ - \ \frac{1}{ \sin \ y } [/tex]

    Getting rid of the minus signs!

    [tex] \frac{dy}{dx} \ = \ \ \frac{1}{ \sin \ y } [/tex]

    Now, from [tex] \sin^2 \ y \ + \ \cos^2 \ y \ = \ 1 [/tex] we solve for sin: [tex] \sin \ y \ = \ \sqrt{1 \ - \ \cos^2 \ y} [/tex]

    [tex] \frac{dy}{dy} \ = \ \frac{1}{ \sqrt{1 \ - \ \cos^2 \ y} } [/tex]

    Now, when we were taking the derivative in the first place we noticed that x = cos y
    so obviously cos²y = x² so we get:

    [tex] \frac{dy}{dx} \ = \ \frac{1}{ \sqrt{1 \ - \ x^2} } [/tex]

    Which was the original integral we began with.
     
  8. Oct 6, 2010 #7
    Re: Trigonometric substitution - Why???

    Wait a minute. Not so fast :-)

    The integration result by your own admission using this method
    "in this manner" is = - arccos x.

    And we know from standard tables and the classic method of
    substituting sin theta for x, the correct result is arcsin x.

    So are you saying that arcsin x is NOT the correct result ?
    Or do you contend that -arccos x = arcsin x ?

    What you did is verify the result by differentiating, thus invoking the
    Fundamental Rule of Calculus ..... Integration and Differentiation are
    inverse functions. But that is irrelevant. The method produces the wrong
    answer depending on how one assigns the sides of the right triangle.

    Are we having fun yet ?
     
  9. Oct 6, 2010 #8
    Re: Trigonometric substitution - Why???

    Haha, forget the tables :wink: Follow the logic, I'll set everything up again.

    [tex] \int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx[/tex]

    Looking at the picture I drew you'll see that we can rewrite the integral as

    [tex] \int \ csc \ \theta \ dx [/tex]

    Now, again if we look at the picture we can express [tex] \cos \ \theta \ = \frac{x}{1}[/tex]

    This is the same as [tex] \cos \ \theta \ = \ x [/tex]. If I take the derivative I get:

    [tex] - \ \sin \ \theta \ d \ \theta \ = \ dx [/tex]

    I can rewrite the integral as:

    [tex] \int \ csc \ \theta \ dx \ = \ - \ \int \ csc \ \theta \ \sin \ \theta \ d \ \theta[/tex]

    We know that [tex] \csc \ \theta \ = \ \frac{1}{ \sin \ \theta } [/tex] so

    [tex] - \ \int \ csc \ \theta \ \sin \ \theta \ d \ \theta \ = \ - \ \int \frac{1}{ \sin \ \theta } \ \sin \ \theta \ d \ \theta \ = \ - \ \int \frac{ \sin \ \theta }{ \sin \ \theta } \ d \ \theta \ = \ - \ \int \ d \ \theta [/tex]

    The final integral is:

    [tex] - \ \int \ d \theta \ = \ - \ \theta \ + \ C[/tex]

    Remember that [tex] \cos \ \theta \ = \ x [/tex]. I can then take the inverse to
    re-express the angle in terms of x's.

    [tex] \cos \ \theta \ = \ x [/tex]

    [tex] \theta \ = \ \arccos \ ( \ x \ )[/tex]

    [tex] - \ \int \ d \theta \ = \ - \ \theta \ = \ - \ arccos \ ( \ x \ ) \ + \ C[/tex]

    Actually, you're going to have to forgive me, at the very end of the last answer I made a mistake.
    When I wrote "x = cos y so obviously cos²y = x²" I forgot
    that I actually set x = - cos y. When I take the derivative of inverse functions I usually go to the side
    of the page and rederive them from basics rather than memorizing them. I should never have
    included the minus when I was deriving the original integral from - arccos x. It kind of invalidated
    everything I wrote in the last response from "Still not convinced?" onwards :tongue:

    One last time! :tongue2:

    Still not convinced? I can use differentiation to check whether I've gotten the right answer
    seeing as integration and differentiation are inverse processes. I should be able to, by
    differentiating, end up with the exact same integral as I started with.

    [tex] y \ = \ arccos \ ( \ x \ ) [/tex]

    [tex] cos \ y \ = \ x \ [/tex]

    [tex] - \ \sin \ y \ \frac{dy}{dx} \ = \ 1 [/tex]

    [tex] - \ \frac{dy}{dx} \ = \ \frac{1}{ \sin \ y } [/tex]

    [tex] \frac{dy}{dx} \ = \ - \ \frac{1}{ \sin \ y } [/tex]

    Now, from [tex] \sin^2 \ y \ + \ \cos^2 \ y \ = \ 1 [/tex] we solve for sin: [tex] \sin \ y \ = \ \sqrt{1 \ - \ \cos^2 \ y} [/tex]

    [tex] \frac{dy}{dy} \ = \ \frac{1}{ \sqrt{1 \ - \ \cos^2 \ y} } [/tex]

    Now, when we were taking the derivative in the first place we noticed that x = cos y
    so obviously cos²y = x² so we get:

    [tex] \frac{dy}{dx} \ = \ - \ \frac{1}{ \sqrt{1 \ - \ x^2} } [/tex]

    This is the derivative of arccosx, if we want the derivative of minus arccos x we just
    negate the sign of this final result and we get

    [tex] \frac{1}{ \sqrt{1 \ - \ x^2} } [/tex]

    Which was the original integral.
     
    Last edited: Oct 6, 2010
  10. Oct 6, 2010 #9
    Re: Trigonometric substitution - Why???

    Yes, I follow that your answer beginning with cos theta = x results in the
    solution to be theta = - arccos x. Your step by step solution looks correct.

    But the problem is that this contradicts
    1/ All integration tables that show the integral to be arcsin x
    2/ The solution you get if you interchange the non hypotenuse sides
    and solve with sin theta = x
    3/ The solution one obtains if using the traditional substitution of
    x = sin theta.

    So I repeat my questions ......
    Are you saying that arcsin x is NOT the correct result ?
    Or do you contend that -arccos x = arcsin x ?

    I do not think there is any interval for a definite integration of this
    integrand which will give the same numerical result for both solutions.

    This is an interesting problem.
     
    Last edited: Oct 6, 2010
  11. Oct 6, 2010 #10
    Re: Trigonometric substitution - Why???

    Testing

    [tex]
    \int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx
    [/tex]

    Thanks for your patience
     
  12. Oct 7, 2010 #11
    Re: Trigonometric substitution - Why???

    Sorry for the late response. Here is a way to test what I've done for yourself.
    Take the definite integral of the function from -0.5 to 0.5 using this triangle method
    I've used but do two integrals. First do the integral the way I did it and evaluate it.
    Then do the integral changing the legs, i.e. set [tex] \sqrt{ 1 \ - \ x^2}[/tex] to be
    the bottom leg. Let me know.
     
  13. Oct 7, 2010 #12
    Re: Trigonometric substitution - Why???

    I think that this discrepancy between the two solutions is accounted for by having different constants of integration.
     
  14. Oct 7, 2010 #13
    Re: Trigonometric substitution - Why???

    If you do the two definite integrals you get the same answer.

    If you do the integral my way you get [tex] \int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx \ = \ - \ cos^{-1}(x) \ + \ C[/tex]

    If you do the integral with the [tex] \sqrt{1 \ - \ x^2[/tex] on the other leg

    you get [tex] \int \ \frac{1}{ \sqrt{1 \ - \ x^2}} \ dx \ = \ sin^{-1}(x) \ + \ C[/tex]

    There is a question of signs as well as the constant of integration at play.
    If you follow the logic accurately you'll get the right answer & you always have
    differentiation to test whether you've gotten the right answer or not.
    For those integrals you can't easily form a right triangle just use some algebra
    like completing the square or factoring etc... to get what you need :wink:
     
  15. Oct 7, 2010 #14
    Re: Trigonometric substitution - Why???

    Well now I can see that the derivatives of sine x and cosine x differ only in sign.
    So that means the integrals of this integrand also differ only in sign.
    So arcsin x does indeed = minus arccos x.

    What is interesting is why interchanging the right triangle sides produces the sign change.

    Note that the exact same thing occurs with arctan x and arccot x
    AND arcsec x and arccsc x. Their derivatives are identical except for sign.

    Why do I love having my brain twisted by Mathematics ?
    I hope its not a masochistic desire ......... hahahahaha

    Cheers
     
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