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Math Help - integration

  1. #1
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    integration

    arccos(2x)dx
    I'm using integration by parts and almost got the right answer
    I got xarccos(2x) - (1/2)radical(1-x^2) + c
    the radical is messed up I think.
    Thanks once again for all the help.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    arccos(2x)dx
    I'm using integration by parts and almost got the right answer
    I got xarccos(2x) - (1/2)radical(1-x^2) + c
    the radical is messed up I think.
    Thanks once again for all the help.
    \int u dv = uv - \int v du

    So for
    \int acs(2x) dx
    use
    u = acs(2x) ==> du = -\frac{2}{\sqrt{1 - 4x^2}}
    and
    dv = dx ==> v = x

    Giving:
    \int acs(2x) dx = x~acs(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}}

    For the integral, let y = 2x ==> dy = 2 dx so...
    \int \frac{2x}{\sqrt{1 - 4x^2}} dx = \frac{1}{2}  \int \frac{y}{\sqrt{1 - y^2}} dy

    Then z = 1 - y^2 ==> dz = -2y dy so
    \int \frac{2x}{\sqrt{1 - 4x^2}} dx = \frac{1}{2} \int \frac{y}{1 - y^2} dy = \frac{1}{2} \cdot -\frac{1}{2} \int \frac{1}{\sqrt{z}} dz  = -\frac{1}{4} \int z^{-1/2} dz = -\frac{1}{4} \cdot \frac{1}{\frac{1}{2}} z^{1/2} = -\frac{1}{2} \sqrt{1 - y^2} = -\frac{1}{2} \sqrt{1 - 4x^2}

    Thus
    \int acs(2x) dx = x~acs(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}} = x acs(2x) - -\frac{1}{2} \sqrt{1 - 4x^2}

    -Dan
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by topsquark View Post
    \int \frac{2x}{\sqrt{1 - 4x^2}}
    Set \mu=\sqrt{1-4x^2}\implies{d}\mu=-\frac{4x}{\sqrt{1-4x^2}}~dx, this yields

     <br />
\begin{aligned}<br />
\int {\frac{{2x}}<br />
{{\sqrt {1 - 4x^2 } }}~dx} &= - \frac{1}<br />
{2}\int {d\mu }\\<br />
&= - \frac{1}<br />
{2}\sqrt {1 - 4x^2 } + k<br />
\end{aligned}<br />
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Set \mu=\sqrt{1-4x^2}\implies{d}\mu=-\frac{4x}{\sqrt{1-4x^2}}~dx, this yields

     <br />
\begin{aligned}<br />
\int {\frac{{2x}}<br />
{{\sqrt {1 - 4x^2 } }}~dx} &= - \frac{1}<br />
{2}\int {d\mu }\\<br />
&= - \frac{1}<br />
{2}\sqrt {1 - 4x^2 } + k<br />
\end{aligned}<br />
    I know this is shorter, but I prefer smaller steps even if it takes me longer. (It helps my poor mind organize things better!)

    -Dan
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Aha... but you know more stuff than me, so what you said does not count
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