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Math Help - [SOLVED] Integration by Parts Help

  1. #1
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    [SOLVED] Integration by Parts Help

    Does anyone know how to do the integral of "4x arcsin(x)dx" ? Help is much appreciated.
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  2. #2
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    Quote Originally Posted by GreySweatshirt306 View Post
    Does anyone know how to do the integral of "4x arcsin(x)dx" ? Help is much appreciated.
    Use integration by parts u = \sin^{-1}x,\;\;\;dv = 4x\,dx
    Last edited by mr fantastic; February 12th 2009 at 04:33 PM. Reason: No edit - just flagging this reply as having been moved from a double posting of the question by the OP
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  3. #3
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    <br />
\begin{gathered}<br />
  I = \int {4x\arcsin (x)dx}  \hfill \\<br />
  u = \arcsin (x) \Rightarrow du = \frac{1}<br />
{{\sqrt {1 - x^2 } }} \wedge dv = x \Rightarrow v = x^2  \hfill \\ <br />
\end{gathered} <br />

    <br />
I = x^2 \arcsin (x) - \int {\frac{{x^2 }}<br />
{{\sqrt {1 - x^2 } }}dx} <br />

    Now

    <br />
\int {\frac{{ - x^2 }}<br />
{{\sqrt {1 - x^2 } }}dx}  = \int {\frac{{1 - x^2  + 1}}<br />
{{\sqrt {1 - x^2 } }}} dx = \int {\sqrt {1 - x^2 } } dx + \int {\frac{1}<br />
{{\sqrt {1 - x^2 } }}dx} <br />

    Can you continue?
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  4. #4
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    Correction:

    \int\frac{-x^2}{\sqrt{1-x^2}}\,dx=\int\frac{1-x^2-1}{\sqrt{1-x^2}}\,dx=\int\sqrt{1-x^2}\,dx-\int\frac{1}{\sqrt{1-x^2}}\,dx.

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  5. #5
    Member Nacho's Avatar
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    Quote Originally Posted by Scott H View Post
    Correction:

    \int\frac{-x^2}{\sqrt{1-x^2}}\,dx=\int\frac{1-x^2-1}{\sqrt{1-x^2}}\,dx=\int\sqrt{1-x^2}\,dx-\int\frac{1}{\sqrt{1-x^2}}\,dx.

    Always I happened me that

    thank you
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