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Thread: Integration by parts

  1. #1
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    Integration by parts

    Hi all,
    Is this method/solution correct?
    thanks
    John

    Integrate by parts;

    $\displaystyle x^2.sin2x $

    $\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $

    $\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

    $\displaystyle v= -\frac{cos2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$

    Substituting

    $\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.dx+C$


    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\-cos2x.dx+C$

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\frac{sin2x}{2}+C$

    $\displaystyle =\frac{-x^2cos2x}{2}+\frac{sin2x}{4}+C$
    Last edited by celtic1234; Sep 11th 2011 at 11:17 AM.
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    Re: Integration by parts

    Quote Originally Posted by celtic1234 View Post
    Hi all,
    Is this method/solution correct?
    thanks
    John

    Integrate by parts;

    $\displaystyle x^2.sin2x $

    $\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $

    $\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

    $\displaystyle v= -\frac{cos2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$

    Substituting

    $\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.dx+C$


    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\-cos2x.dx+C$

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\frac{sin2x}{2}+C$

    $\displaystyle =\frac{-x^2cos2x}{2}+\frac{sin2x}{4}+C$
    You have the rule wrong. The rule is $\displaystyle \displaystyle \int{u\,dv} = u\,v - \int{v\,du}$ (or if you prefer the longer notation: $\displaystyle \displaystyle \int{u\,\frac{dv}{dx}\,dx} = u\,v - \int{v\,\frac{du}{dx}\,dx}$ ).
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  3. #3
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    Re: Integration by parts

    sorry
    this is my proper solution;
    can you tell me if i am following the correct method-is the solution correct?
    thanks
    John

    Integrate by parts;

    $\displaystyle x^2.sin2x $

    $\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $

    $\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

    $\displaystyle v= -\frac{cos2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$

    Substituting

    $\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.dx+C$


    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\-cos2x.dx.2x+C$

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\frac{sin2x}{2}.2x+C$

    $\displaystyle =\frac{-x^2cos2x}{2}+\frac{2xsin2x}{4}+C$
    Last edited by celtic1234; Sep 11th 2011 at 12:53 PM.
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  4. #4
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    Re: Integration by parts

    Quote Originally Posted by celtic1234 View Post
    sorry
    this is my proper solution;
    can you tell me if i am following the correct method-is the solution correct?
    thanks
    John

    Integrate by parts;

    $\displaystyle x^2.sin2x $

    $\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $
    Good so far.
    $\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

    $\displaystyle v= -\frac{cos2x}{2}+C $
    Absolutely fine to here.
    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int\frac{du}{dx}.dx+C$
    And then you stray slightly. The rule is, simply put and imprecisely:

    $\displaystyle \int u\dfrac{dv}{dx}=u\cdot{v}-\int v\dfrac{du}{dx}$

    When I substitute, this becomes:


    $\displaystyle \int x^2sin2x=x^2\cdot{-\frac{cos2x}{2}}-\int -\frac{cos2x}{2}\cdot 2x~ dx$

    $\displaystyle \int x^2sin2x=-x^2\cdot{\frac{cos2x}{2}}+\int x\cdot cos2x~dx$

    And I think you'll have to use parts again here, making this a particularly bothersome question.
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    Re: Integration by parts

    Thanks
    Is this it?

    Integrate by parts;

    $\displaystyle x^2.sin2x $

    $\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $

    $\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

    $\displaystyle v= -\frac{cos2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$


    Substituting

    $\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C$

    Need to solve the integral of 2x.Cos2x by parts:

    $\displaystyle \int 2x.cos2x $

    $\displaystyle let \ u=2x, then\ \frac {du}{dx}=2.dx $

    $\displaystyle let \ \frac{dv}{dx}=cos2x, then\ v= \int (cos2x).dx $

    $\displaystyle v= \frac{sin2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

    Substitute back in;

    $\displaystyle =2x.\frac{sin2x}{2}-\int\frac{sin2x}{2}.2.dx+C$

    $\displaystyle =2x.\frac{sin2x}{2}-\frac{2}{2}\int sin2x.dx+C$

    $\displaystyle =2x.\frac{sin2x}{2}-1.\frac{-cos2x}{2}+C$

    $\displaystyle =2x.\frac{sin2x}{2}+\frac{cos2x}{2}+C$

    $\displaystyle =x.sin2x+\frac{cos2x}{2}+C$
    Last edited by celtic1234; Sep 12th 2011 at 10:53 AM.
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  6. #6
    Super Member Quacky's Avatar
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    Re: Integration by parts

    Quote Originally Posted by Quacky View Post
    And then you stray slightly. The rule is, simply put and imprecisely:

    $\displaystyle \int u\dfrac{dv}{dx}=u\cdot{v}-\int v\dfrac{du}{dx}$

    When I substitute, this becomes:


    $\displaystyle \int x^2sin2x=x^2\cdot{-\frac{cos2x}{2}}-\int -\frac{cos2x}{2}\cdot 2x~ dx$

    $\displaystyle \int x^2sin2x=-x^2\cdot{\frac{cos2x}{2}}+\int x\cdot cos2x~dx$

    And I think you'll have to use parts again here, making this a particularly bothersome question.
    Quote Originally Posted by celtic1234 View Post
    Thanks
    Is this it?

    Integrate by parts;

    $\displaystyle x^2.sin2x $

    $\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $

    $\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

    $\displaystyle v= -\frac{cos2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$


    Substituting

    $\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C$

    Need to solve the integral of 2x.Cos2x by parts:

    $\displaystyle \int 2x.cos2x $

    $\displaystyle let \ u=2x, then\ \frac {du}{dx}=2.dx $

    $\displaystyle let \ \frac{dv}{dx}=cos2x, then\ v= \int (cos2x).dx $

    $\displaystyle v= \frac{sin2x}{2}+C $
    Good so far.

    $\displaystyle rule=\ \int u \frac{du}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

    Substitute back in;

    $\displaystyle =2x.\frac{sin2x}{2}-\int\frac{sin2x}{2}.2.dx+C$
    Yep
    $\displaystyle =2x.\frac{sin2x}{2}-\frac{2}{2}\int sin2x.dx+C$

    $\displaystyle =2x.\frac{sin2x}{2}-1.\frac{-cos2x}{2}+C$

    $\displaystyle =2x.\frac{sin2x}{2}+\frac{cos2x}{2}+C$

    $\displaystyle =x.sin2x+\frac{cos2x}{2}+C$
    Yes, I think that's fine!
    You made it difficult for yourself. When you get to this stage:

    $\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

    You can cancel the 2s and rewrite, as I had:

    $\displaystyle =-x^2\cdot{\frac{cos2x}{2}}+\int x\cdot cos2x~dx$

    Which is easier to work with.

    Anyway, what's the answer to your original question, which was:

    $\displaystyle \int x^2sin(2x)~dx$ ? You don't have to work anything else out. Use these two facts:

    $\displaystyle \int x^2sin(2x)~dx=x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C$

    $\displaystyle \int\2x.cos2x~dx=x.sin2x+\frac{cos2x}{2}$
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    Re: Integration by parts

    Your notation is also off.

    If $\displaystyle \displaystyle u = x^2$ then EITHER $\displaystyle \displaystyle \frac{du}{dx} = 2x$ OR $\displaystyle \displaystyle du = 2x\,dx$, NOT $\displaystyle \displaystyle \frac{du}{dx} = 2x\,dx$.
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    Re: Integration by parts

    Thanks all
    this is my full solution:
    Sorry for reposting it all but in case anyone else wanted to follow the full solution-(i am sure to confuse them otherwise)
    Your time is appreciated
    Hopefully i did not make anymore mistakes!
    thanks
    John

    Integrate by parts;

    $\displaystyle x^2.sin2x $

    $\displaystyle let \ u=x^2, then\ \frac {du}{dx}=2x.dx $

    $\displaystyle let \ \frac{dv}{dx}=sin2x, then\ v= \int (sin2x).dx $

    $\displaystyle v= -\frac{cos2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{dv}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$


    Substituting

    $\displaystyle =x^2.\frac{-cos2x}{2}-\int\frac{-cos2x}{2}.2x.dx+C$

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C.......equa tion (1)$

    Need to solve the integral of 2x.Cos2x by parts:

    $\displaystyle \int 2x.cos2x $

    $\displaystyle let \ u=2x, then\ du=2.dx $

    $\displaystyle let \ \frac{dv}{dx}=cos2x, then\ v= \int (cos2x).dx $

    $\displaystyle v= \frac{sin2x}{2}+C $

    $\displaystyle rule=\ \int u \frac{dv}{dx}.dx=u.v-\int v\frac{du}{dx}.dx+C$

    Substitute back in;

    $\displaystyle =2x.\frac{sin2x}{2}-\int\frac{sin2x}{2}.2.dx+C$

    $\displaystyle =2x.\frac{sin2x}{2}-\frac{2}{2}\int sin2x.dx+C$

    $\displaystyle =2x.\frac{sin2x}{2}-1.\frac{-cos2x}{2}+C$

    $\displaystyle =2x.\frac{sin2x}{2}+\frac{cos2x}{2}+C$

    $\displaystyle =x.sin2x+\frac{cos2x}{2}+C$

    Now substitute this integral back into equation (1)

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}\int\2x.cos2x.dx+C.......equa tion (1)$

    $\displaystyle =x^2.\frac{-cos2x}{2}+\frac{1}{2}[x.sin2x+\frac{cos2x}{2}]$

    $\displaystyle =-x^2.\frac{cos2x}{2}+\frac{x.sin2x}{2}+\frac{cos2x} {4}$
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  9. #9
    Super Member Quacky's Avatar
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    Re: Integration by parts

    Nice! Perfect solution.
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    Re: Integration by parts

    thanks Quacky
    appreciate the help
    I am working on another one now that has an Lnx term in it-it seems to be repeating itself over and over-no thought there is a simple solution -i may well have to post about it though....keep the eyes peeled!
    John
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    Re: Integration by parts

    no doubt...not no thought? i am getting worse by the day...
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    Re: Integration by parts

    Okay

    Maybe this will help - for $\displaystyle \int ln(x)~dx$, let $\displaystyle \frac{dv}{dx}=1$ and $\displaystyle u=ln(x)$
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    Re: Integration by parts

    Hi Quacky,
    If the function is

    $\displaystyle x^3.lnx $

    it repeats and repeats.

    Now if i make

    $\displaystyle u= x^3 and \frac{dv}{dx}=lnx $

    Then i need to integrate the $\displaystyle lnx$ term to get v.

    however if integrate it by parts and make

    $\displaystyle \ u=1 $ and$\displaystyle \frac{dv}{dx}=lnx $

    then it repeats over and over.

    However if i make:

    $\displaystyle u=lnx$ and$\displaystyle \frac{dv}{dx}=1 $

    then i can get a solution fairly easily.

    Is it ok to rearrange the $\displaystyle lnx $ and the $\displaystyle 1.dx$ to suit ?

    Do i need to post this as a different thread?

    regards
    John
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    Re: Integration by parts

    Sorry to butt in, but, just in case a picture helps...

    Quote Originally Posted by celtic1234 View Post
    Hi Quacky,
    If the function is

    $\displaystyle x^3.lnx $

    it repeats and repeats.
    Not so.


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

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,



    ... is lazy integration by parts, doing without u and v.



    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; Sep 13th 2011 at 10:00 AM.
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  15. #15
    Super Member Quacky's Avatar
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    Re: Integration by parts

    I don't know if you followed that, but let $\displaystyle u=ln(x)$ because you can differentiate that to give $\displaystyle \frac{1}{x}$ which simplifies the further stages simply.

    Let $\displaystyle \frac{dv}{dx}=x^3$

    Follow this approach for all $\displaystyle ln(x)$ type questions.
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