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Math Help - Tricky Integral

  1. #1
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    Tricky Integral

    \int 1/(7-5sin(x))\ dx

    What is method of this? Many helps
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    With the substitution t=\tan (x/2) we get a rational integral on t .
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  3. #3
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    Thanks for respond

    How is apply?

    What I can do is by partial fraction? Sorry what I need to substitute because we do not have a t in the question?
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  4. #4
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    Answer to tricky integral

    Hi,
    Indeed this is a tricky integral that can be solved via substitution.

    Steps

    • \int{\dfrac{1}{7-5\sin(x)}}dx
    • Substitution

      u=\tan\dfrac{x}{2}.

      Then, du=\dfrac{1}{2}\sec^2(\dfrac{x}{2})dx

      And, \sin(x)=\dfrac{2u}{u^2+1}, dx=\dfrac{2du}{u^2+1}
    • \int{\dfrac{2}{(u^2+1)\left( 7-\dfrac{10u}{u^2+1} \right) }}du
    • With simplification, you will get:

      2\int{\dfrac{1}{7u^2-10u+7}}
    • By completing the square, you will get:

      2\int \dfrac{1}{\left( \sqrt{7}u-\dfrac{5}{\sqrt{7}} \right)^2+\dfrac{24}{7}}du
    • Another substitution:

      s=\sqrt{7}u-\dfrac{5}{\sqrt{7}}

      ds=\sqrt{7}du
    • You will get:

      \dfrac{2}{\sqrt{7}} \int \dfrac{1}{s^2+\dfrac{24}{7}}
    • The answer of this integral is:

      \dfrac{\arctan\left({\frac{1}{2}\sqrt{\frac{7}{6}}  s}\right)}{\sqrt{6}}+C

      Where C is the integration constant.
    • You then go all the way back to substitute back the elements you used. For instance, you will plug into the resulting integral s=\sqrt{7}u-\frac{5}{\sqrt{7}} so that you get a function of u. You will then substitute u for u=\tan{\left(\dfrac{x}{2}\right)}


    The answer is:

    -\dfrac{\arctan{\left(\dfrac{5-7\tan{\frac{x}{2}}}{2\sqrt{6}}\right)}}{\sqrt{6}}+  C
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  5. #5
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    You magnificent to help this much I can offer 100 thanks but it only let me 1. I much pleased. I will not realise these step myself. But I hope one day I can help many people as you will help me. This is what I am much pleased for. You are gracious kindness much.

    Much apology but how is sin(x) found to be as to be 2u/u^2+1?
    Last edited by Caption12; April 3rd 2011 at 10:04 AM. Reason: Add question
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  6. #6
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    \displaystyle{\eq \sin x\ =\ 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})\ =\ 2 \tan(\frac{x}{2}) \cos^2(\frac{x}{2})\ =\ \frac{2 \tan(\frac{x}{2})}{\sec^2(\frac{x}{2})}\ =\ \frac{2 \tan(\frac{x}{2})}{1 + \tan^2(\frac{x}{2})}}
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  7. #7
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    Thank to you balloons, this is much inteligent to know this!

    I have one more ask, it is how you are doing this integral to get arctan form

    As I know d/dx arctan x = 1/x^2+1, but what here we have is not one it is 1/s^2 + 24/7

    now i ask how this 24/7 will change the way you must to work on this step?

    hope understand is clear what i ask

    thankyou (i will work "24/7" and wont find this solution without the helps :P)
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  8. #8
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    \displaystyle{\frac{1}{s^2 + \frac{24}{7}}\ =\ \frac{1}{\frac{24}{7}(\frac{7}{24}s^2 + 1)}}

    \displaystyle{\ =\ \frac{7}{24} \frac{1}{(\sqrt{\frac{7}{24}}s)^2 + 1}}

    \displaystyle{\ =\ \frac{7}{24} \frac{1}{(\frac{1}{2}\sqrt{\frac{7}{6}}s)^2 + 1}}}
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  9. #9
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    no sorry that does not to help me, but much thank, i know you want help me,

    i dont know how it relate to arctangent direvative.

    as far i know only d/dx arctanx=1/x^2 + 1

    i dont know it with other number but x^2 and 1, i dont know it when you have the co - eficient of x^2 or with another number than one

    i know i would normal use my chainrules, but this not posible here

    can you tell me is this also

    d/dx arctan fx = 1/(fx)^2 + 1

    can you only use this with a ONE and not another number

    is it only posible, if the number term is made to 1

    i ask reason is this, up to know, i have not see arctangent direvative. this is day 1 i try this piece... and is advance question.
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  10. #10
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    Just in case a picture helps...



    ... 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 s), 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...


    Hope this helps to see how the chain rule is indeed the key. Others may post a more conventional working with Leibniz notation, but this way works for me.

    NB of course I've only dealt with the 's' part of the journey.
    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  11. #11
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    I know the meaning from the balon, but it will not answer what i ask..

    as i say i do not know the direvative arctan FX i can oonly if is just artan X. so i especialy dont kknow the rules for this. i especaily dont know to work it backward..

    how u can integrate the 1/(x^2 + 24/7) i dont know this.

    can some one please say, i will be very appreciate, what is definition direative of arctan fx

    also you have put as constant rt7/2rt6

    but should i be (2/rt7)*(7/24) equals as rt7/12
    Last edited by Caption12; April 3rd 2011 at 06:34 PM.
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Caption12 View Post
    I know the meaning from the balon, but it will not answer what i ask..

    as i say i do not know the direvative arctan FX i can oonly if is just artan X. so i especialy dont kknow the rules for this. i especaily dont know to work it backward..

    how u can integrate the 1/(x^2 + 24/7) i dont know this.
    AS tom@ballooncalculus has already shown you:
    \displaystyle \int \frac{1}{x^2 + \frac{24}{7}}~dx

    \displaystyle = \frac{7}{24} \int \frac{1}{\left ( \frac{7}{24} \right ) x^2 + 1}~dx

    Now let y = \sqrt{\frac{7}{24}} x \implies x = \sqrt{\frac{24}{7}} y \implies dx = \sqrt{\frac{24}{7}}~dy, so the integral becomes:

    \displaystyle = \frac{7}{24} \int \frac{1}{y^2 + 1} \cdot \sqrt{\frac{24}{7}}~dy

    \displaystyle = \frac{7}{24} \cdot \sqrt{\frac{24}{7}} \int \frac{1}{y^2 + 1}~dy

    -Dan
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  13. #13
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    that is not what i have ask

    the change of x to y you post dan is not helpful, in fact, is not the method used earlier which i spend time understanding, you actually made thing much harder and unecesary, and out subsitution is x so term of y will not help

    what i ask before (I CAN work it out if i know this... but no one answer this) is WHAT is direvative what arctan fx?

    not arctan x but arctan fx how is chain rule use here when result is 1/x^2 + 1 because how it will affect on to the 1 in denom. i ask for general rule because i cannot find this from searching internet and is not in text book..


    ALSO

    when u are reaching the final answer

    why is it u are putting

    -arctan(5-7tan) instead of arctan(7tan-5)? why is the need to change this aspect becaue it is extrawork for the same event?

    anyway the final step is to evaluate between limits 0 and 2pi?

    please help, i make the definite integral ZERO because the tan(pi) and tan(0) are both zero the two terms cancel. how can i make it beetter
    Last edited by Caption12; April 3rd 2011 at 10:41 PM. Reason: `
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  14. #14
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    Quote Originally Posted by Caption12 View Post
    also you have put as constant rt7/2rt6

    but should i be (2/rt7)*(7/24) equals as rt7/12
    Well, I was deliberately ignoring the given constant of 2/rt7 and focussing on 1/(s^2 +24/7), but unfortunately I did mis-write that starting point (in the bottom left of the picture) as 1/(s^2 + 1), and this might well have confused you (or anybody) trying to make sense of the integration from the picture (now corrected). Try again working anti-clockwise from bottom left, noting that we're replacing

    \displaystyle{\frac{1}{s^2 + \frac{24}{7}}}

    right away with

    \displaystyle{\sqrt{\frac{7}{24}}\ \frac{1}{(\sqrt{\frac{7}{24}}\ s)^2 + 1}\ \sqrt{\frac{7}{24}}}

    which is the same as the re-arrangement given previously except for splitting 7/24 into root (7/24) times root (7/24).

    Then note that we can replace the inner function [root(7/24) s] with tan theta, which results in an inner derivative of sec^2 theta, from which we can deduce that the top (integral) function F is arctan.

    Now, you very sensibly want to be able to make sense of all this by looking at differentiating arctan of a function, using the chain rule. I was of course hoping that you would see the advantage of the diagram in doing just this - i.e. following it clockwise from top left instead of anti-clockwise from bottom left. But maybe my typo threw you off. Anyway, you will find plenty of examples of differentiating arctan of an inner function, but I hope you will look specifically at my similar examples at here and here.
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  15. #15
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    thankyou much i understand how to get to the result now

    but please some body emergency help me with this definite intigral limits

    0 to 2pi

    tom baloon you have confuse me a litle bit with the tan theta parts but what you have wirte in post #8 i can follow more well... i can use chainrule from this

    but i need someone tell me is this now corrct

    d/dx arctan fx will equal as (f'x)1/((fx)^2 + 1)

    because what i did to to compare by this result which fernando post is divide the coficient of the integral by direvative of INSIDe this bracket (fx=(1/2)sqrt(7/6)x)

    and i DID come to result which i will understand if and only if this general result i have assumed is actualy true?

    tom i can folow your picture i undertaand what you have to doing in this but i dont know myself the rules formulas you have used and i want to apply it myself without just copy up the answer in to my study book... i want next time is not to have ask
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