Results 1 to 13 of 13

Math Help - Integration question relating to inverse trig functions.

  1. #1
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4

    Integration question relating to inverse trig functions.

    Hi,

    I've been doing some differentiation of inverse trig. In doing this I had to find the derivative of:

    <br />
arctan (3x/2)

    I found this to be: 6/(9x^2 + 4)

    To check this answer I thought I would try to integrate to see if I arrived back at the original question. However, I'm a bit stumped as to how I should go about integrating it?

    I know that the original derivative was derived using the chain rule so is there some substitution that I can use to find the integral of 6/(9x^2 + 4)

    Thanks,
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    Your question asked for the derivative of
    y = arctan(3x/2)

    so x = 2tan(y)/3

    Use this substitution for the integral. It should simplify very nicely (assuming you did your differentiation correctly).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Commit this to memory \displaystyle  \int \frac{a}{a^2+x^2}~dx = \text{Tan}^{-1}\frac{x}{a}+C
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4
    Thanks for the quick reply.

    How would you find the right substitution if you were just given the question:

    Integrate 6/9x^2 +4

    I'm sure that I could spot that it related to arctan (the x^2 and the positive constant is the give away) but how could I get it into a form from which I could integrate?

    Sorry if I'm missing the obvious...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4
    Got it now.

    If I compare  6/9x^2 +4

    With

    \displaystyle  \int \frac{a}{a^2+x^2}~dx = \text{Tan}^{-1}\frac{x}{a}+C

    Divide through by 9 to make the coeff of x^2 equal to 1. Immediately you have the general form given above.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    It is actually standard to try to substitute trig functions when you see forms like ax^2 + b in the denominator of an integral (without an x term in the numerator).
    Normally you use the trig function that allows you to simplify the expression with a trig identity.
    For example when you see 1 - x^2 in the denominator, you might use x = sin(y).
    There may be multiple trig subs (or even hyperbolic subs, i.e. sinh, cosh) that work, but they should all come out to be equivalent answers.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4
    I can see how you could reasonably easily work out which trig or hyperbolic to sub in. But given the integration question above, all I would have been able to tell you was that I should sub in tan. I wouldn't have been able to get the rest of the substitution at all...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    Sure you can the 9 and 4 have something to do with the 3 and 2

    Normally, if you have (a + bx^2) you think a(1 + \frac{b}{a}x^2).
    Now the coefficient should cancel out \frac{b}{a}.
    Last edited by snowtea; January 4th 2011 at 06:40 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    You can also find the result on your own if you don't remember the derivative/identity

    \displaystyle \int \frac{6}{9x^2+4}~dx

    Make \displaystyle x = \frac{2}{3}\tan\theta \implies \theta = \tan^{-1}\frac{3x}{2} and \displaystyle dx=\frac{2}{3}\sec^{2}\theta d\theta

    \displaystyle \int \frac{6}{9( \frac{2}{3}\tan\theta )^2+4}\times \frac{2}{3}\sec^{2}\theta d\theta

    \displaystyle \int \frac{6}{9\times \frac{4}{9}\tan^2\theta +4}\times\frac{2}{3}\sec^{2}\theta d\theta

    \displaystyle \int \frac{6}{4\tan^2\theta +4}\times\frac{2}{3}\sec^{2}\theta d\theta

    \displaystyle \int \frac{6}{4(\tan^2\theta +1)}\times\frac{2}{3}\sec^{2}\theta d\theta

    \displaystyle \int \frac{6}{4\sec^2\theta }\times\frac{2}{3}\sec^{2}\theta d\theta

    Cancelling down

    \displaystyle \int 1 ~d\theta  = \theta +C  = \tan^{-1}\frac{3x}{2}+C
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4
    Thanks for all the replies. They've been extremely helpful.

    My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?

    I've tried just subing in x = tan y. In the above example as a test to see if it's really always necessary to include in the correct coefficients and I haven't managed to make it work yet. (which may well be because my algebra is somewhat lacking...)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by Ant View Post
    My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?
    No, but with enough practice, you'll be able to spot the relevant substitution (at least for the regular ones).
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Ant View Post
    My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?
    Practice makes perfect here, the more you do the better you will get at identifying which substitution to use. Trial and error won't hurt, patterns will present.

    Look here for more info

    Trigonometric substitutions

    Quote Originally Posted by Ant View Post

    I've tried just subing in x = tan y. In the above example as a test to see if it's really always necessary to include in the correct coefficients and I haven't managed to make it work yet. (which may well be because my algebra is somewhat lacking...)
    Maybe this is the case, once again practice will make perfect, you are looking to cancel any coeffecients leaving something neat like 1-\cos^2\theta, 1-\sin^2\theta , 1+\tan\theta
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Quote Originally Posted by Ant View Post
    My only niggling problem is that in future I may be given slightly more complicated question in which the necessary substitution isn't quite so obvious. I think I'll always be able to identify which trig or hyperbolic I should use but is there a general way to work out what the correct full sub should be?

    I've tried just subing in x = tan y. In the above example as a test to see if it's really always necessary to include in the correct coefficients and I haven't managed to make it work yet.
    Instead of assuming you need a sub in the form

    x = some function of tan theta,

    you might prefer to get the denominator in the form 1 + some square or other, thereby making it analogous to the left hand side of 1 + \tan^2\theta = \sec^2\theta, which is the whole point of the sub. Then it's obvious to sub tan theta for directly for the square you made.

    Just in case a picture helps...



    Then adjust for this expression being a derivative via the chain rule



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case ), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    The general drift is...



    Then you can make the direct analogy with integrating 1 + tan squared...



    See also http://www.mathhelpforum.com/math-he...tml#post546765

    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question relating to solving equations with Green's functions
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: March 9th 2012, 09:23 PM
  2. Replies: 3
    Last Post: January 20th 2011, 09:02 AM
  3. Integration Question (inverse trig)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 10th 2010, 06:53 PM
  4. Inverse Trig Functions and Advance Trig
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 24th 2008, 03:13 PM
  5. Question Regarding Inverse Trig Functions
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: September 6th 2008, 07:43 PM

Search Tags


/mathhelpforum @mathhelpforum