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A Smart Way to Avoid a Lot of Inverse Hassle?

  1. May 31, 2010 #1
    Hi everybody :approve:

    I'm having a lot of trouble remembering all of the formulas for differentiating inverse trig functions, for integrating trig functions, or integrating inverse trig functions, or integrating hyperbolic trig functions, or integrating inverse hyperbolic trig functions.

    Luckily, in differentiating inverse trig functions I can invert the equation & use implicit differentiation & rederive all 6 trig functions all just by knowing how to differentiate trig functions.

    That to me is a lot smarter & better than cheating by memorizing formulas as there is simple logic involved.

    However, I've had so much trouble trying to deal with the rest of these functions in a logical way & every source I check tells me to just use the formula's I've memorized.

    Thomas Calculus
    Stewart Calculus
    Apostol Calculus
    Calculus Made Easy
    The Calculus Lifesaver
    Loads of youtube videos
    Loads of internet pdf's and sites

    I've had some success with a particular triangle method, i.e. look for the sum of two perfect squares in a denominator under a square root in the denominator, that means you can draw a triangle & use a very similar logic to rederive your way to the answer according to the situation.

    Namely the logic purported in this and this.

    However, what about all these seperate cases,

    inverse trig functions,
    hyperbolic trig
    inverse hyperbolic trig

    Have all of you just memorized the formula's to just plug in every time you recognise the shape of the equation?

    If not I'd be so happy to hear of sources to where you learned this smart way that I seem to be missing. I've had so much trouble with calculus in that I have to ignore my books & search out the real secret behind each method in calculus, for instance;

    Newtons method is just an algebraic manipulation of the point-slope form of an equation, two seconds to rederive it.
    Linear Approximation is another manipulation of this simple equation in a slightly different way.
    Mean Value Theorem is just Rolle's Theorem slanted.
    Integral MVT is just standard averaging calc'd up.
    All of that shells and washers stuff is just geometry & a flavour of calc.

    There just has to be some way to look at trig integrals in a similarly logical way.
  2. jcsd
  3. May 31, 2010 #2


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    Good for you! That is a nice summary and you are doing exactly the right thing to help you understand and remember the concepts and formulas.

    One minor nit-pick. The MVT is sort of like a rotated Rolle's theorem, but not exactly. The reason it is never proved that way is that expressing the function f in terms of rotated coordinates may make it no longer a single valued function. You might like the argument that considers (a,f(a)), (b,f(b)), and (x,f(x)) for x in [a,b] as the three vertices of a triangle. Then the area of the triangle is proportional to

    [tex]F(x)= \left |\begin{array}{ccc}

    1 &a& f(a)\\1& b& f(b)\\1& x &f(x)
    \end{array}\right |

    F(a) and F(b) are obviously 0 and applying Rolle's theorem to F gives the MVT.

    I don't think you will find a nice way to group stuff like integrating the inverse trig functions.

    P.S. And welcome to the forums.
  4. May 31, 2010 #3
    That's actually great thinking of it as a rotated triangle that varies from (a,f(a)) to (b,f(b)).

    The proof I know uses, again, the point-slope formula to reduce the whole thing to Rolle's theorem subtracting the secant connecting (a,f(a)) to (b,f(b)) from f(x).

    But, I'm really curious. How do you deal with the 12 inverse trig/hyperbolic functions you may come across?

    I've been thinking & I think it's possible to try to break up the hyperbolic stuff into it's e^x components but the inverse stuff is out of this world.

    As far as standard inverse trig functions go, if you check those links for one second you'll see what I mean by re-deriving everything using a triangle & I'm very concerned about the failings of this method. I've had success with it so far but I'm afraid to blankly rely on it without checking with people who know. I also wonder what people do since I haven't read about this method anywhere except those videos.

    Also, I don't see how they relate to inverse hyperbolic trig at all & how to intuitively understand these bad boys...

    Thanks for reading and helping :biggrin:
  5. May 31, 2010 #4


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    Are you and gerardholye the same person?

    I wouldn't be too concerned if I were you. About the only advice I can give about the hyperbolic functions is to note the similarity of their derivative formulas to the ordinary trig functions. And, as far as trying to integrate them, just remember that the inverse functions are easier to differentiate than integrate. So, for example if I want to do this integral:

    [tex]\int \cosh^{-1}(x)\, dx[/tex]

    I would try integration by parts letting u = cosh-1(x), dv = 1dx

    This gets rid of the inverse function at the expense of an extra x. Try it; you'll like it.
  6. May 31, 2010 #5
    Yeah my firefox & laptop has been going crazy l8ly so I just created a new seperate account on a lot of the sites I'm on.

    Alright, I'll look out for little things like that & mess around & be back if anything grabs my attention.

    Thanks a lot :biggrin:
  7. May 31, 2010 #6
    The extra x is somewhat helpful actually.
    x=sinht works.
    It still involves int by parts (ew) but at least it's obvious what to do and it's sort of a "brain off, pen on" integral.
  8. Jun 3, 2010 #7
    Here is an example of the method I use that works pretty well, it's got a lot of kinks I'm trying to iron out, and you'll see what I mean.

    [PLAIN]http://img684.imageshack.us/img684/6945/trianglef.jpg [Broken]

    [itex]\int \frac{dx}{\sqrt{e^{2x} - 6}} [/itex]

    by the picture,

    [itex] tan \theta = \frac{\sqrt{6}}{\sqrt{e^{2x} - 6}} [/itex]


    [itex] \frac{1}{\sqrt{6}}tan \theta = \frac{1}{\sqrt{e^{2x} - 6}} [/itex]

    so the integral becomes;

    [itex] \int \frac{dx}{\sqrt{e^{2x} - 6}} =\int \frac{1}{\sqrt{6}}tan \theta dx [/itex]

    Now to calculate dx, (this gets a bit nuts);

    [itex] u = e^x [/itex]

    [itex] du = e^x dx [/itex]

    [itex] dx = \frac{du}{e^x} = \frac{du}{u} [/itex]

    so, the integral is NOW;

    [itex] \int \frac{1}{\sqrt{6}} \ tan \theta \ \frac{du}{u} [/itex]

    We first get rid of the [itex]u[/itex] by using the fact that,

    [itex] sin \theta = \frac{\sqrt{6}}{u} [/itex]

    [itex] \frac{1}{\sqrt{6}} \ sin \theta = \frac{1}{u} [/itex]

    and this gets our integral even closer;

    [itex] \int \frac{1}{\sqrt{6}} \ tan \theta \ \frac{1}{\sqrt{6}} \ sin \theta du [/itex]

    cleaning it up;

    [itex] \int \frac{1}{6} \ tan \theta \ sin \theta du [/itex]

    All thats left to get rid of is [itex]du[/itex],


    [itex] sin \theta = \frac{\sqrt{6}}{e^x} [/itex]


    [itex] e^x = u = \frac{\sqrt{6}}{sin \theta} = \sqrt{6} \ csc \theta [/itex]

    we see that;

    [itex] u = \sqrt{6} \ csc \theta [/itex]


    [itex] du = - \sqrt{6} \ csc \theta \ cot \theta \ d \theta [/itex]

    and the original integral finally becomes:

    [itex] \int \frac{1}{6} \ tan \theta \ sin \theta \ du = \int - \frac{1}{6} \ tan \theta \ sin \theta \ \sqrt{6} \ csc \theta \ cot \theta \ d \theta [/itex]

    Cleaning it up;

    [itex] \int - \frac{1}{\sqrt{6}} \ tan \theta \ sin \theta \ csc \theta \ cot \theta \ d \theta [/itex]

    and this reduces to;

    [itex] - \frac{1}{\sqrt{6}} \int d \theta [/itex]

    [itex] - \frac{1}{\sqrt{6}} \ \theta + C [/itex]

    to get theta I'll use;

    [itex] sin \theta = \frac{\sqrt{6}}{e^x} [/itex]

    [itex] \theta = arcsin ( \frac{\sqrt{6}}{e^x}) [/itex]

    so the final answer I get is;

    [itex] - \frac{1}{\sqrt{6}} \ arcsin ( \frac{\sqrt{6}}{e^x}) + C [/itex]


    [itex] \int \frac{dx}{\sqrt{e^{2x} - 6}} dx = - \frac{1}{\sqrt{6}} arcsin( \frac{\sqrt{6}}{e^x}) + C [/itex]

    But my book says the answer is;

    [itex] \int \frac{dx}{\sqrt{e^{2x} - 6}} = \frac{1}{\sqrt{6}} arcsec( \frac{e^x}{\sqrt{6}}) + C [/itex]

    I am supposed to blatantly copy backwards using the method my book
    teaches me to do but there is no real magic in this method, it is contrived magic with no thinking and no nothing to copying.

    There must of been some slip-up but I haven't seen it :eek:

    Could you correct me in this method, it is very general insofar as I can tell,
    there might be some completing the square in the denominator to get to this
    form but it's very general insofar as I can tell.

    What did I do wrong here?

    I don't think I'm differentiating right as I keep getting weird answers.

    But, the method I've used, where is it wrong?
    Last edited by a moderator: May 4, 2017
  9. Jun 3, 2010 #8
    "cheating by memorizing formulas"-???
    I should have called my prof a cheater when the quadratic formula came up in the course if solving PDE's!!!
  10. Jun 3, 2010 #9

    I'm sorry but in memorizing the quadratic formula are there 6 very similar variations that, if remembered incorrectly, will lead to a completely wrong answer?

    Also, are there 6 Very similar inverse hyperbolic representations of the quadratic equation that are very similar & if confused when recalling from memory will lead to a completely wrong answer?

    I can do a side calculation and compute a heckload of derivatives if I'm stuck and then reverse copy if things get hectic but that's not very satisfying & while some people are happy to memorize these formula's I doubt everyone does & I bet there's a more logical way.

    If you're into a PDE's class I'm sure you could help me really easily with this question, and I really doubt you have these 12 integrals memorized after taking in so much from further classes.
  11. Jun 5, 2010 #10

    I'll just reply to the original post rather than the recent nonsense.

    Drawing the triangles is the most intuitively accessible route for me. You might be looking for a "simplification" that doesn't exist. In "the real world" (whatever THAT means!), arbitrary equations don't just present themselves.

    One's ability/willingness to memorize formulas (and/or their derivations) or mnemonic devices just depends on the focus of their studies.
  12. Jun 5, 2010 #11
    This is the books answer differentiated;

    [itex] y = \frac{1}{\sqrt{6}} \cdot arcsec(\frac{e^x}{\sqrt{6}}) + C [/itex]

    [itex] sec(\sqrt{6}y) = sec[arcsec(\frac{e^x}{\sqrt{6}})] + sec(C) [/itex]

    [itex] sec(\sqrt{6}y) = \frac{e^x}{\sqrt{6}} + sec(C) [/itex]

    [itex] sec(\sqrt{6}y) tan(\sqrt{6}y) \frac{dy}{dx} = \frac{e^x}{\sqrt{6}} [/itex]

    [itex] \frac{dy}{dx} = \frac{e^x}{\sqrt{6}} \cdot \frac{1}{ sec(\sqrt{6}y) tan(\sqrt{6}y) } [/itex]

    [itex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \cdot \frac{e^x}{\sqrt{6}} \cdot \frac{1}{ tan(\sqrt{6}y) } [/itex]

    [itex] \frac{dy}{dx} = \frac{1}{ tan(\sqrt{6}y) } [/itex]

    [itex] tan(\sqrt{6}y) = \sqrt{sec^2(\sqrt{6}y) - 1} [/itex]

    [itex] \frac{dy}{dx} = \frac{1}{ \sqrt{sec^2(\sqrt{6}y) - 1} } [/itex]

    [itex] \frac{dy}{dx} = \frac{1}{ \sqrt{\frac{e^{2x}}{6} - 1} } [/itex]

    I'm not sure exactly how to legally factor that in the denominator to get it into the form the original integral is in but I'm 99.9999% sure it is legal. Could you help?

    This is my answer differentiated;

    [itex] y = - \frac{1}{\sqrt{6}} \cdot arcsin ( \frac{\sqrt{6}}{e^x}) + C [/itex]

    [itex] sin(\sqrt{6}y) = - \frac{\sqrt{6}}{e^x} + sin(C) [/itex]

    [itex] cos(\sqrt{6}y) \cdot \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} [/itex]

    [itex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \cdot \frac{1}{cos(\sqrt{6}y)} [/itex]

    [itex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \cdot \frac{1}{\sqrt{1 - sin^2 (\sqrt{6}y)}} [/itex]

    [itex] \frac{dy}{dx} = \frac{\sqrt{6}}{e^x} \cdot \frac{1}{\sqrt{1 - \frac{6}{e^{2x}}}} [/itex]

    [STRIKE] The next terms wont show in latex for some reason but they
    are just;
    cos(root6 y) becomes the square root of 1 - sin^2(root6 y) and then converted into
    6 over e^2x
    [/STRIKE] (Forget about this, It worked when I clicked enter ;)

    Again, I have a feeling that this can be factored into the correct form
    as the original integral but I don't see how to do it legally.

    However, there is the case of that extra term sticking out like a sore thumb!

    If this answer can be factored then I've gotten it amost right, apart from that term which I maybe missed :confused:

    I have a feeling this is correct in a similar way that sinθ = -cosθ' what with the argument inside the arcsin and the arcsecant being flipped as well as the inclusion of the minus sign, would that be correct even in arcsin land & is there a surefire way to discern that naturally?
    Last edited: Jun 5, 2010
  13. Jun 5, 2010 #12
    On "theirs" -
    \frac{dy}{dx} = \frac{1}{ \sqrt{\frac{e^{2x}}{6} - 1} }

    Multiply top and bottom by sqrt(6)/e^(2x) and it should reconcile with yours.
  14. Jun 5, 2010 #13


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    I didn't read through your method, but it looks like you get the same answer as your book. You just need to note a few things: First find how arcsin and arccsc are related; and then find a relationship between arcsin and arccos. After doing this, you should see that your answer is correct.
  15. Jun 6, 2010 #14
    Hmm... I checked wikipedia and found these identities;

    [itex] arcsec(\frac{1}{x}) = arccos(x) [/itex]

    [itex] arccos(x) = \frac{\pi}{2} - arcsin(x) [/itex]

    [itex]arcsec(\frac{1}{x}) = \frac{\pi}{2} - arcsin(x) [/itex]


    that answers the invertability question & I would wager that the [tex]\frac{\pi}{2} [/tex] is taken care of by the constant & if evaluating this as a definite integral it would work itself out.

    Would that be right?

    I don't think I got the differentiation right because that extra term does not dissappear if I multiply by 1 on top and bottom like chaz says, idk...
  16. Jun 6, 2010 #15
    cos-1 x = y
    x = cos y
    1/x = 1/cos y = sec y
    sec-1(1/x) = y

    cos-1 x = sec-1(1/x)

    Using the identity cos-1 x + sin-1 x = [itex]\pi[/itex]/2,
    -sin-1 x = cos-1 x - [itex]\pi[/itex]/2 = sec-1(1/x) - [itex]\pi[/itex]/2
    and yes, the -[itex]\pi[/itex]/2 gets absorbed in the constant of integration C. That's how you get the difference between the two with the negative sign and reciprocal arguments in the inverse functions.
  17. Jun 6, 2010 #16
    The system works!

    After a week on this problem, the system is validated :!!)

    Thanks so much guys/girls :biggrin:

    Now, to go on and make hyperbolic functions work like this :eek:

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