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Math Help - difficult antiderivative

  1. #1
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    difficult antiderivative

    Hi,

    Can someone please explain to me how to find the antiderivative of x^3/3(cosmx)? I'm trying to solve the interval of x^2cosmx dx and I am at the point where I need to find the antiderivative of x^3/3(cosmx). I don't have any idea how to solve this. I would know how to solve it if I had to find the derivative, but I don't know what to do here since its the antiderivative.

    Thank you very much
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi,

    Can someone please explain to me how to find the antiderivative of x^3/3(cosmx)? I'm trying to solve the interval of x^2cosmx dx and I am at the point where I need to find the antiderivative of x^3/3(cosmx). I don't have any idea how to solve this. I would know how to solve it if I had to find the derivative, but I don't know what to do here since its the antiderivative.

    Thank you very much
    do integration by parts. use the polynomial part as the part to differentiate and the trigonometric part as the part to integrate all the time (you will have to do it 3 times or so). there's probably a formula for an integral of this type, i don't do well with memorizing formulas so i can't tell you
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  3. #3
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    Hi,

    I did it four times and it looks worse now. Does anyone see what I did wrong?

    This is what I did:

    Integral x^2cosmx dx

    u=cosmx
    dv=x^2dx

    du=-sinmx
    v=x^3/3

    integral x^2cosmx dx=(cosmx)(x^3/3)-integral(x^3/3(cosmx)

    then I did it again:

    u=x^3
    dv=cosmx dx

    du=2x^2/3 (quotient rule)
    v=sinmx

    (x^3)/3(sinmx)-integral(sinmx)(2x^2/3)

    so...(cosmx)(x^3/3)-(x^3/3)(sinmx)-integral sin(mx)(2x^2/3)

    then I did it again

    u=2x^2/3
    dv=sinmx

    du=4/3x
    v=-cosmx

    (cosmx)(x^3/3)-(x^3)/3(sinmx)-(2x^2/3)(-cosmx)-integral (-cosmx)4/3x

    then I simplified

    (cosmx)(x^3/3)-(x^3/3)(sinmx)+(2x^2/3)cosmx-integral -cosmx(4/3x)

    then I did it again

    u=2x^2/3
    dv=cosmx

    du=12x+2x^2/3(-sinmx)/9 (this doesn't look like its going to help)

    v=sinmx

    Thank you very much
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  4. #4
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    \int x^2\cos(mx)\,dx=\frac{2x\cos(mx)}{m^2}+\frac{(m^2x  ^2-2)\sin (mx)}{m^3}+k

    First, set y=mx, after that apply integration by parts twice.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi,

    I did it four times and it looks worse now. Does anyone see what I did wrong?

    This is what I did:

    Integral x^2cosmx dx

    u=cosmx
    dv=x^2dx

    du=-sinmx
    v=x^3/3

    integral x^2cosmx dx=(cosmx)(x^3/3)-integral(x^3/3(cosmx)

    then I did it again:

    u=x^3
    dv=cosmx dx

    du=2x^2/3 (quotient rule)
    v=sinmx

    (x^3)/3(sinmx)-integral(sinmx)(2x^2/3)

    so...(cosmx)(x^3/3)-(x^3/3)(sinmx)-integral sin(mx)(2x^2/3)

    then I did it again

    u=2x^2/3
    dv=sinmx

    du=4/3x
    v=-cosmx

    (cosmx)(x^3/3)-(x^3)/3(sinmx)-(2x^2/3)(-cosmx)-integral (-cosmx)4/3x

    then I simplified

    (cosmx)(x^3/3)-(x^3/3)(sinmx)+(2x^2/3)cosmx-integral -cosmx(4/3x)

    then I did it again

    u=2x^2/3
    dv=cosmx

    du=12x+2x^2/3(-sinmx)/9 (this doesn't look like its going to help)

    v=sinmx

    Thank you very much
    i thought we were finding \frac {1}{3} \int x^3 \cos mx~dx , but now i see that it was \int x^2 \cos mx~dx. in that case, i suggest what Krizalid does. in any case, you didn't do what i said, which is why you ended up in trouble. i said differentiate the polynomial and integrate the trig function ("always"), you did the opposite. you will keep getting a worst looking integral if you continue doing it the way you're doing it. switch u and dv
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  6. #6
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    We have to do it the long way Does this look right to you guys?

    Integral x^2cosmxdx

    u=x^2
    dv=cosmxdx

    du=2x
    v=sinmx

    integral x^2cosmx=x^2sinmx-integral sinmx(2x)

    u=2x
    dv=sinmx

    du=2
    v=-cosmx

    integral x^2cosmx=x^2sinmx-[(2x)(-cosmx)] -integral(-cosmx(2))
    =x^2sinmx+2xcosmx+ integral (-cosmx(2)
    x^2sinmx+2xcosmx+ integral(-1)cosmx(2)

    u=2
    dv=-cosmx dx

    du=0
    v=-sinmx

    =x^2(sinmx)+2xcosmx-integral (cosmx)(2)
    =x^2sinmx+2xcosmx-[2(-sinmx)]-integral -sinx(0)
    =x^2sinmx+2xcosmx+2sinmx

    Thank you very much
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    We have to do it the long way Does this look right to you guys?

    Integral x^2cosmxdx

    u=x^2
    dv=cosmxdx

    du=2x
    v=sinmx

    integral x^2cosmx=x^2sinmx-integral sinmx(2x)

    u=2x
    dv=sinmx

    du=2
    v=-cosmx

    integral x^2cosmx=x^2sinmx-[(2x)(-cosmx)] -integral(-cosmx(2))
    =x^2sinmx+2xcosmx+ integral (-cosmx(2)
    x^2sinmx+2xcosmx+ integral(-1)cosmx(2)

    u=2
    dv=-cosmx dx

    du=0
    v=-sinmx

    =x^2(sinmx)+2xcosmx-integral (cosmx)(2)
    =x^2sinmx+2xcosmx-[2(-sinmx)]-integral -sinx(0)
    =x^2sinmx+2xcosmx+2sinmx

    Thank you very much
    did your professor say you have to do it the long way?

    this is incorrect by the way. you took the integrals wrong. the red line above is your first mistake, there are others, but you have to fix the first one. look out for that mistake again. you have to use substitution for the integral, you don't have to show it, and it's not that hard to do in your head
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  8. #8
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    Hey chocolatelover

    On integration by parts, always we have to make a "little integration"

    Since dv=\cos(mx)\,dx, then v=\frac1m\sin(mx) (the constant is omitted on the integration by parts), why?

    Simple, imagine you have to integrate \int\cos(mx)\,dx, so first, let's set a change of variables according to u=mx\implies du=m\,dx\implies dx=\frac1m\,du, so now we substitute and yields

    \int\cos(mx)\,dx=\frac1m\int\cos u\,du, and this is easy to integrate, 'cause it's a basic integral, so it yields \frac1m\int\cos u\,du=\frac1m\sin u+k, finally back substitute u into mx and we happily get \int\cos(mx)\,dx=\frac1m\sin(mx)+k

    Is it clearer now?
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  9. #9
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    Hi Krizalid and Jhevon,

    Sorry. I'm still kind of confused. I tried to do it agian.

    I let u=x^2
    dv=cosmxdx
    du=2x
    v=sinmx

    Then I said that X^2cosmxdx=x^2sinmx-integral sinmx(2x)

    Then wouldn't I do a u subtitution?

    However, I don't know what to let u equal. If I let u equal sinmx then du would be cosmx, and that wouldn't work. If I let u equal 2x then du would be 2 and that wouldn't work. My professor told us that we had to do it the long way. Do you see what I'm doing wrong?

    Thank you very much
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  10. #10
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    Quote Originally Posted by chocolatelover View Post
    dv=cosmxdx
    v=sinmx
    Didn't you see my previous post?
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  11. #11
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    Yes, but I don't understand it. Sorry. Is that the first step and then do you integrate it by using the integration by parts method?
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  12. #12
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    Would this work?

    u=x^2
    dv=cosmx dx

    du=2x
    v=sinmx

    the integral x^2cosmx dx=x^2(sinmx)-the integral (sinmx)(2x)

    the integral sinmx(2x)=2x(-cosmx)-the integral (-2cosmx)

    u=2x
    dv=sinmx

    du=2
    v=-cosmx

    the integralx^2cosmxdx=x^2sinmx-2x(-cosmx)-the integral(-2cosmx)

    the integral (-2cosmx
    u=-2
    dv=cosmx
    du=0
    v=sinmx

    the integral x^2cosmxdx=x^2sinmx-2x(-cosmx)-2(sinmx-integral(sinmx)(0)=
    x^2sinmx+2xcosmx-2sinmx+c

    Thank you
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Would this work?

    u=x^2
    dv=cosmx dx

    du=2x
    v=sinmx

    the integral x^2cosmx dx=x^2(sinmx)-the integral (sinmx)(2x)

    the integral sinmx(2x)=2x(-cosmx)-the integral (-2cosmx)

    u=2x
    dv=sinmx

    du=2
    v=-cosmx

    the integralx^2cosmxdx=x^2sinmx-2x(-cosmx)-the integral(-2cosmx)

    the integral (-2cosmx
    u=-2
    dv=cosmx
    du=0
    v=sinmx

    the integral x^2cosmxdx=x^2sinmx-2x(-cosmx)-2(sinmx-integral(sinmx)(0)=
    x^2sinmx+2xcosmx-2sinmx+c

    Thank you
    again, you keep doing what we said you're not supposed to do. the integral of cos(mx) IS NOT sin(mx) and the integral of sin(mx) IS NOT -cos(mx)

    By substitution (see Krizalid's post):
    \int \sin mx ~dx = - \frac {1}{m} \cos mx + C and \int \cos mx ~dx = \frac {1}{m} \sin mx + C

    until you get the pieces right, you will never get the whole solution right
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  14. #14
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    Okay, let's do this.

    \int x^2\cos(mx)\,dx

    Set y=mx\implies dy=m\,dx, then the integral becomes to

    \int x^2\cos(mx)\,dx=\frac1{m^3}\int y^2\cos y\,dy

    Now, we're gonna integrate y^2\cos y, so we'll use integration by parts twice.

    Let u=y^2\implies du=2y\,dy & dv=\cos y\,dy\implies v=\sin y, the integral becomes to

    \int y^2\cos y\,dy=y^2\sin y-2\underbrace{\int y\sin y\,dy}_I

    Apply integration by parts again on I

    Let u=y\implies du=dy & dv=\sin y\,dy\implies v=-\cos y, this yields

    I=-y\cos y+\int\cos y\,dy=-y\cos y+\sin y (we'll add the constant at the end of the integration)

    Back substitute

    \int y^2\cos y\,dy=y^2\sin y-2(\sin y-y\cos y), then

    \int x^2\cos(mx)\,dx=\frac1{m^3}[m^2x^2\sin(mx)-2(\sin(mx)-mx\cos(mx))]+k

    Which is equal to

    \int x^2\cos(mx)\,dx=\frac1{m^3}[2mx\cos(mx)+(m^2x^2-2)\sin(mx)]+k; now multiply by \frac1{m^3} and we are done.

    By the way, I suggest you brush up some of derivatives, it's quite dangerous get start on integration without a fully domain of derivatives.
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  15. #15
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    I understood of all that except, how did you get 1/m^3? I understand where you got 1/m from, but how did it change to m^3? Also, at the end of the problem you said to multiply it by 1/m^3. Wouldn't you either leave it how it is or multiply by m^3 on both sides?

    Thank you very much

    Regards
    Last edited by chocolatelover; September 1st 2007 at 06:34 PM.
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