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Math Help - Solving integrals by parts :D

  1. #1
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    Cool Solving integrals by parts :D

    Hi, I'm having a little trouble solving an integral by parts, here is what I have so far:

    The integral of xsin(x/2)dx

    I let u(random letter) = x.
    so du = dx.
    and dv = sin(x/2).
    and v = cos(x/2).

    Then I do UV-the integral of dvu.

    I then get:

    xCos(x/2) - the integral of sin(x/2)x dx

    And this is the part I get confused on, am I supposed to do:
    xCos(x/2) - -Cos(x/2)x dx
    which means
    xCos(x/2) + Cos(x/2)x dx
    And from here I am lost, does anyone know what I should do?
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  2. #2
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    Quote Originally Posted by vandorin View Post
    Hi, I'm having a little trouble solving an integral by parts, here is what I have so far:

    The integral of xsin(x/2)dx

    I let u(random letter) = x.
    so du = dx.
    and dv = sin(x/2).
    and v = cos(x/2).

    Then I do UV-the integral of dvu.

    I then get:

    xCos(x/2) - the integral of sin(x/2)x dx

    And this is the part I get confused on, am I supposed to do:
    xCos(x/2) - -Cos(x/2)x dx
    which means
    xCos(x/2) + Cos(x/2)x dx
    And from here I am lost, does anyone know what I should do?
    1. Your "v" is wrong.
    2. Its better to use the substitution t=\frac{x}{2} First. then use integration by parts.
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  3. #3
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    Quote Originally Posted by vandorin View Post
    Hi, I'm having a little trouble solving an integral by parts, here is what I have so far:

    The integral of xsin(x/2)dx

    I let u(random letter) = x.
    so du = dx.
    and dv = sin(x/2).
    and v = cos(x/2).

    Then I do UV-the integral of dvu.

    I then get:

    xCos(x/2) - the integral of sin(x/2)x dx

    And this is the part I get confused on, am I supposed to do:
    xCos(x/2) - -Cos(x/2)x dx
    which means
    xCos(x/2) + Cos(x/2)x dx
    And from here I am lost, does anyone know what I should do?
    You have the wrong function for V

    V=-2cos\left(\frac{1}{2}x\right)

    You can check that by differentiating. Always make sure you have the right functions before you apply IBP.

    Now you just need to apply the formula

    \int udv=uv-\int vdu

    =-2xcos\left(\frac{1}{2}x\right)-\int -2cos\left(\frac{1}{2}x\right)dx
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  4. #4
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    Quote Originally Posted by adkinsjr View Post
    You have the wrong function for V

    V=-2cos\left(\frac{1}{2}x\right)

    Now you just need to apply the formula

    \int udv=uv-\int vdu

    =-2xcos\left(\frac{1}{2}x\right)-\int -2cos\left(\frac{1}{2}x\right)dx

    OH! Is v=-2xCos because v is the antiderivative of dv? so it is what you have to take the derivative of to get dv?

    But I am confused as to why there is a -2 infront of the second cos?
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  5. #5
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    Quote Originally Posted by vandorin View Post
    OH! Is v=-2xCos because v is the antiderivative of dv? so it is what you have to take the derivative of to get dv?

    But I am confused as to why there is a -2 infront of the second cos?
    Correct. To find the "v", you will integrate "dv".
    To find your "v", Evaluate \int sin(\frac{x}{2}) dx=\int sin(\frac{1}{2}x)dx
    The result of this integral is your "v". (Take the constant of integration = 0).
    To evaluate this integral, substitute t=\frac{1}{2}x.
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  6. #6
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    Quote Originally Posted by vandorin View Post
    OH! Is v=-2xCos because v is the antiderivative of dv? yes, but v=2cos, and u=x. The isn't an x infront of the cosine until we combine u and v in the formula \int udv=uv-\int vdu

    so it is what you have to take the derivative of to get dv?

    But I am confused as to why there is a -2 infront of the second cos?
    Remember that the derivative of [cos(x)=-sin(x) That's where the negative sign comes from. So if

    v=-2cos\left(\frac{1}{2}x\right)

    \frac{dv}{dx}=sin\left(\frac{1}{2}x\right)
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  7. #7
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    Quote Originally Posted by General View Post
    Correct. To find the "v", you will integrate "dv".
    To find your "v", Evaluate \int sin(\frac{x}{2}) dx=\int sin(\frac{1}{2}x)dx
    The result of this integral is your "v". (Take the constant of integration = 0).
    To evaluate this integral, substitute t=\frac{1}{2}x.
    Ok, thanks . I understand why my v was wrong now. I still don't understand why there would be a 2 in front of the second cos though? because I'm multiplying u*v - the integral of sin(1/2 x) x

    So how does that come out to be -2cos etc? And plus my book is telling me the answer is something like
    -2xCos(x/2)+4Sin(x/2)+c

    I'm completely lost on everything except how to get the c haha.
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  8. #8
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    Quote Originally Posted by vandorin View Post
    Ok, thanks . I understand why my v was wrong now. I still don't understand why there would be a 2 in front of the second cos though? because I'm multiplying u*v - the integral of sin(1/2 x) x

    So how does that come out to be -2cos etc? And plus my book is telling me the answer is something like
    -2xCos(x/2)+4Sin(x/2)+c

    I'm completely lost on everything except how to get the c haha.
    I told you : Your "v" = \int sin(\frac{1}{2}x) dx , and take the constant of the integration of this integral = 0.

    You can not integrate this simple integral?
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  9. #9
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    Quote Originally Posted by General View Post
    I told you : Your "v" = \int sin(\frac{1}{2}x) dx , and take the constant of the integration of this integral = 0.

    You can not integrate this simple integral?
    I can, and so the whole equation would look like:

    -2xCos(1/2 x) - -2Cos(1/2x)x Because the integral of sin is -cos, but won't this equation come out to just be x? since the -2Cos(1/2 x) and +2Cos(1/2x) cancel out, so you are just left with x?
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  10. #10
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    Quote Originally Posted by vandorin View Post
    I can, and so the whole equation would look like:

    -2xCos(1/2 x) - -2Cos(1/2x)x Because the integral of sin is -cos, but won't this equation come out to just be x? since the -2Cos(1/2 x) and +2Cos(1/2x) cancel out, so you are just left with x?
    No, Your integration for the \int v du is wrong.
    If you apply the integration by parts for the original integral with u=x and dv=sin(\frac{x}{2}) dx, you will face:
    -2xcos(\frac{x}{2}) + 2 \int cos(\frac{x}{2})dx.
    Now, integrate the new integral again. And be careful when you do that.
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  11. #11
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    Quote Originally Posted by General View Post
    No, Your integration for the \int v du is wrong.
    If you apply the integration by parts for the original integral with u=x and dv=sin(\frac{x}{2}) dx, you will face:
    -2xcos(\frac{x}{2}) + 2 \int cos(\frac{x}{2})dx.
    Now, integrate the new integral again. And be careful when you do that.
    Ah, I've got it now. I was thinking that the formula was "UV-DVU" When it is actually "UV-VDU" That is what was throwing me off. Thanks for the help!
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