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Integration Problem

  1. Nov 21, 2003 #1
    I have been working on this integration problem for awhile now and am completely stuck:

    Integral of tanx*tan2x*tan3x

    I know I am suppose to use substituion and for that matter probably multiple substitions. But I am no sure what I should use for my first value of "u". I have changed the problem to be in terms of sin and cos, but I am still just ending up at a dead end. If some one can help me start off in the right step I would greatly appreciatte it. I DO NOT want the answer, I would like to solve it for myself, I just need help starting it off right. Thanks in advance.
     
  2. jcsd
  3. Nov 21, 2003 #2
    How about changing the form of your integrand using trig identities before attempting to integrate?

    i.e.: tan(x+y) = ?
     
  4. Nov 21, 2003 #3
    tan(x+y) = (tanx +tany)/(1-tanxtany) right?

    But I dont see how that can help me. :-\
     
  5. Nov 21, 2003 #4
    ...

    Try using the expansion of tan (x+y) to simplify tan 2x and tan 3x and try to bring all the three terms in terms of tanx only. Now substitute tanx=sinx/cosx and simplify the expression you get to see the below answer:

    I = the integral = I1+I2;
    where,
    I1 = int[(4sin^3xcosx)dx]/(cos^2x-Sin^x) and

    I2 = int[(2Sin^3xSecx)dx]/(cos^2x-Sin^x)

    Put cosxdx = d(sinx) and then put, sinx = t and then I1 reduces to

    int(4x^3dx/1-2X^2), solve this integral (can u???). Repeat a similar procedure for I2 and lo behold u have the answer with u...


    Sridhar
     
  6. Nov 21, 2003 #5
    tan(2x) = tan(x + x)
    tan(3x) = tan((x+x) + x)

    So you end up with an expression in powers of tan(x), no tan(2x) and no tan(3x).

    Then, integrate.
     
    Last edited: Nov 21, 2003
  7. Nov 21, 2003 #6
    I'm sorry but I'm just not gettin how you did that. How did you expand the integral of tan(x)*tan(2x)*tan(3x) to the integral you gave? Is there some trig identity I am not aware of?
     
  8. Nov 21, 2003 #7
    Ok, I get that alittle better now, I'll start working on it like that and see where I get, thanks guys, Ill post again if I get lost again, which I probably will.
     
  9. Nov 22, 2003 #8
    Ok I've been working on it for awhile now after changing the problem to:

    integral tan(x)*tan(x+x)*tan(x+x+x)

    I don't see how that is making the equation more favorable for me to solve. I still have to use substitution in the next step. I am just really confused at this point and am very desperate to get this done.
     
  10. Nov 22, 2003 #9
    ...

    Your question has already been answered...Just scroll up the page a li'l bit and u have the answer there. I have given the step after using tan(a+b) and then the step after substituting tanx = sinx/cosx and then the method to solve the final integral...

    I think u understand that cosx dx = d(sinx) {as d(sinx)/dx = cosx}
    similarly, sinxdx = -d(cosx); These substitutions will help u solve the final integral that I have given u... That is what I have explained previously....


    If u still do not understand follow these procedures:

    1) Use tan(a+b) to bring the entire integral in terms of tanx alone. Note you might have to use tan2x = 2tanx/sec^2x.

    2)After you have got an expression in terms of tanx:

    integral{[(4tan^3x)/(sec^2x(Sec^2x-2tan^2x))]+ [(2tan^3x)/(sec^2x-2tan^2x)]}

    3) now, substitute, tanx = sinx/cosx, you will get:

    I (the main integral) = I1 + I2
    where,

    I1 = integral{[(4sin^3xcosxdx)/(cos^2x-Sin^2x)]}
    I2 = integral{[(2sin^3xsecx)/(cos^2x-sin^2x)]}

    4) Now consider I1 and I2 seperately and solve them seperately in the following way:

    (i)To solve I1 - cos^2x = 1 - sin^2x and cosxdx = d(sinx). Now substitute sinx = t to get :

    I1 = integral{[(4t^3)dt/(1-2t^2)]}. Can u solve this??? {to solve this put 1-2t^2 = y so that -4tdt = dy and then proceed}

    (ii) To solve I2 - Proceed in a similar way as done for I1 to get:

    I2 = integral{[(1-cos^2x)(sinxdx)/((cosx)(2cos^2x-1))]}
    substitute cosx = p and sinxdx = -d(cosx) to get:

    I2 = integral{[(1-p^2)dp/(p(2p^2-1))]} (u can solve this by adding and subtracting 2x^2 from the numerator to get 2 terms, u can easily solve the 2 terms to get I2)

    5) Add the solved values of I1 and I2 to get the value of integral(tanxtan2xtan3x)...


    Got it???????



    Sridhar
     
  11. Nov 22, 2003 #10
    How does tan2x = 2tanx/sec^2x? I do not see how this can be the same. They produce different y values when you use an x. Is there some kind of trig identitiy that states this? Once I understand this I will be bale to do the problem.
     
  12. Nov 22, 2003 #11

    Hurkyl

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    He got the tangent identity wrong. It is

    [tex]
    \tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}
    [/tex]

    so

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


    However, a general rule to use when confused about trig identities is to convert everything into sines and cosines. (reason: sines and cosines are, in general, much easier to manipulate) If you're doing an integral, you also want to convert everything to having the same argument, if possible. (reason: you're almost always going to have to do this anyways, so you might as well do it up front so it's easier to see the proper substitution to make) Applying this rule would have you do the conversion as follows:

    [tex]
    \tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x}
    [/tex]


    If all else fails, make the substitution

    [tex]
    z = \tan \frac{x}{2}
    [/tex]

    Though this approach will tend to be far messier. You can show, using trig identities, that when using this substitution:

    [tex]\sin x = \frac{2z}{1+z^2}[/tex]
    [tex]\cos x = \frac{1-z^2}{1+z^2}[/tex]
    [tex]dx = \frac{dz}{1+z^2}[/tex]

    So this will convert any rational function of trig functions into an ordinary (but possibly very ugly) rational function which you can do via partial fractions.
     
    Last edited: Nov 22, 2003
  13. Nov 22, 2003 #12
    ...

    Oops Sorry, As hurkyl said, I made a mistake in the identity. However the steps are the same. Bring every term to their sine and cosine forms and then proceed with the problem.

    Sridhar
     
  14. Nov 22, 2003 #13
    Ok, it makes more sense now. I really hate trig identities! I will most likely work on it tomorrow and I should be able to get an answer now. Thanks alot for your help guys, I really appreciatte it!
     
  15. Nov 23, 2003 #14
    Ok a couple more things: First, how did you get those values of sinx, cosx, and dx from z=tan(x/2)? Secondly, I know the integral will be (sinx/cosx)(2sin^2(x)cos(x)/cos^3(x)-sin^2(x)*tan3x. But what does the tan3x equal in terms of sin and cos? I did some work and got confused again as to what value I should use for substituion. Sorry for being such a pain with this, but I do truely appreciatte you guys helping me out.
     
  16. Nov 23, 2003 #15

    Hurkyl

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    You have a parenthesis error.


    From the integral table in my CRC handbook. :smile:

    They're not hard to do by hand, though. E.G.

    [tex]
    \begin{equation*}
    \begin{split}
    \sin x &= \sin \left( 2 \frac{x}{2} \right) \\
    &= 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \\
    &= 2 \frac{z}{\sqrt{1+z^2}} \frac{1}{\sqrt{1+z^2}} \\
    &= \frac{2z}{1+z^2}
    \end{split}
    \end{equation*}
    [/tex]

    (If you don't see how to go from line 2 to line 3, try drawing a triangle with angle [tex]x/2[/tex]; one such triangle has an opposite side with length [tex]z[/tex], and an adjacent side with length [tex]1[/tex]. Note that the square root could be either positive or negative, but since it's being multiplied by itself, it doesn't matter)


    Well, start by writing it in terms of sin and cos!

    [tex]
    \tan 3x = \frac{\sin 3x}{\cos 3x} = \frac{\sin (x + 2x)}{\cos(x + 2x)} = \ldots
    [/tex]

    Can you post all of the work you did, so we can check to make sure it's right, and to show you the way to go about figuring out the next step?
     
    Last edited: Nov 23, 2003
  17. Nov 23, 2003 #16
    Ok, Ill try that. My work is wrong becasue I was using the wrong thing for tan3x. I had:

    The integral of (sinx/cosx)(2sinxcosx/cos^2x-sin^2X) * the integral of (sinx/cosx)(2sinxcosx/cos^2x-sin^2X)

    So then I distibuted the sinx and cosx and got:

    The integral of 2sin^2xcosx/cos^3x-sin^2xcosx * The integral of 2sin^2xcosx/cos^3x-sin^2xcosx

    So I then attempted to use substituion. I used a couple terms and did not get far at all. With all those trig function to the Nth powers I didn't know how to find the derivaitives. That is something we never learned.
     
  18. Nov 23, 2003 #17

    Hurkyl

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    How did you get the product of two integrals?


    Derivatives of powers of trig functions is easy; use the chain rule... why do you think such derivatives are necessary?

    [tex]
    \frac{d}{dx}\left( (\sin x)^n \right) = n (\sin x)^{n-1} \cos x
    [/tex]


    Finally, you need to use parentheses better; 2sinxcosx/cos^2x-sin^2X means

    [tex]
    \frac{2 \sin x \cos x}{\cos^2 x} - \sin^2 x
    [/tex]

    But 2 sin x cos x / (cos^2 x - sin^2 x) means

    [tex]
    \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x}
    [/tex]
     
  19. Nov 25, 2003 #18
    [tex]\int tan(x)tan(2x)tan(3x) dx = \int tan(x)tan(2x)tan(2x + x) dx[/tex]
    [tex]=\int \frac{sin(x)}{cos(x)}(\frac{2tan(x)}{1-tan^2(x)})(\frac{tan(2x) + tan(x)}{1 - tan(2x)tan(x)}) dx[/tex]
    [tex]=\int \frac{sin(x)}{cos(x)}(\frac{sin(2x)}{cos(2x)})(...) dx[/tex]

    u = -cosx
    du = sinx dx.

    [tex]=\int \frac{sin(x)}{-u}(\frac{2sin(x)(-u)}{u^2 - (1 + u^2)})(...) \frac{du}{sin(x)}[/tex]
    [tex]=\int \frac{1}{-u}(u - \frac{1}{3u})(2u) du[/tex]
    [tex]= -2\int u - \frac{1}{3u} du[/tex]
    [tex]= \frac{2ln|-cosx|}{3} - cos^2(x) + C[/tex]

    Not sure if I got it right or not...
     
  20. Nov 29, 2003 #19
    Ok I understand where your going, but you lost me on this step:
    [tex]=\int \frac{sin(x)}{-u}(\frac{2sin(x)(-u)}{u^2 - (1 + u^2)})(...) \frac{du}{sin(x)}[/tex]
    to this step:
    [tex]=\int \frac{1}{-u}(u - \frac{1}{3u})(2u) du[/tex]

    Plus, how did your make the cos(2x) into (u^2)-(1+u^2)?
     
  21. Nov 29, 2003 #20

    Hurkyl

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    [tex]
    \begin{equation*}
    \begin{split}
    I &= \int \tan(x) \tan(2x) \tan(3x) \, dx

    \\ &=
    \int \frac{\sin(x)}{\cos(x)} \frac{\sin(2x)}{\cos(2x)} \frac{\sin(3x)}{\cos(3x)} \, dx

    \\ &= \int \frac{(\sin x) (2 \sin x \cos x) (\sin x (4 \cos^2 x - 1))}
    {(\cos x) (2 \cos^2 x - 1) (4 \cos^3 x - 3 \cos x)} \, dx

    \\ &= \int
    \frac{2 (\sin^3 x) (\cos x) (4 \cos^2 x - 1)}
    {(\cos^2 x) (2 \cos^2 x - 1) (4 \cos^2 x - 3)} \, dx

    \\ &= 2 \int
    \frac{(1 - \cos^2 x) (\cos x) (4 \cos^2 x - 1)}
    {(\cos^2 x) (2 \cos^2 x - 1) (4 \cos^2 x - 3)} (\sin x \, dx)

    \\ &= 2 \int
    \frac{(1 - u^2) u (4 u^2 - 1)}{u^2 (2 u^2 - 1) (4 u^2 - 3)} du

    \\ &= \ldots

    \end{split}
    \end{equation*}
    [/tex]
     
    Last edited: Nov 29, 2003
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