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Math Help - Integration By Parts Help

  1. #1
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    Question Integration By Parts Help

    Hey Guys,

    Can someone please explain to me why the constant isn't used in the v of the u
    v part of the formula?

    I.e. Integral u
    dv = integral uv - integral vdu

    For example Integral x cos (x) dx

    Let x = u and dv = cos(x)dx, then du = 1 and v = sin(x)+c1
    => Integral x cos (x)dx = x sin(x)+c1 - Integral sin(x)+c1 dx
    = x sin(x)+c1 - (Integral sin(x)dx + Integral c
    1 dx)
    = x sin(x)+c1 - (cos(x) + c2 + c1x +c2))
    = x sin(x)+c1 + cos(x) - c2 - c1x - c2

    => x sin(x) +c1 + cos(x) - c2 - c1x - c2 = x sin(x) +c1 + cos(x) - c1x - 2c2 .

    Now the generic solution is xsin(x) + cos(x) + c.

    What am I doing wrong?

    Thanks.

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  2. #2
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    Re: Integration By Parts Help

    Quote Originally Posted by Foreverlearning View Post
    Can someone please explain to me why the constant isn't used in the v of the u●v part of the formula?
    I.e. Integral u●dv = integral u●v - integral v●du
    Because it is not necessary. We use $+c$ only to say that any constant may work.
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  3. #3
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    Re: Integration By Parts Help

    Thanks for the fast reply,

    I realise that it is not necessary but could you please show me mathematically why this requirement is not necessary?

    Could you please show how the long-winded mathematical calculation is equivalent to the shorter methods values?

    I am having trouble wrapping my head around how two different constant values and one of those values also multiplying the variable equates to the same value.

    I look forward to your reply.
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  4. #4
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    Re: Integration By Parts Help

    Quote Originally Posted by Foreverlearning View Post
    I realise that it is not necessary but could you please show me mathematically why this requirement is not necessary?
    Could you please show how the long-winded mathematical calculation is equivalent to the shorter methods values?
    I am having trouble wrapping my head around how two different constant values and one of those values also multiplying the variable equates to the same value.
    Consider the two functions $y(x)=x\sin(x)+\pi~\&~z(x)=x\sin(x)-\pi$. Clearly they are different functions.
    But they have the same derivative, $y'(x)=z'(x)$

    So the anti-derivative of $\sin(x)+x\cos(x)$ is unique up to a constant: $x\sin(x)+c$
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  5. #5
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    Re: Integration By Parts Help

    Thankyou.

    That helps a lot.
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