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Math Help - integral question..

  1. #1
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    integral question..

    \int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx

    u=x u'=1
    v'=\frac{1}{\sin ^2x}dx


    how to solve integral of arc cot??
    Last edited by mr fantastic; March 11th 2009 at 05:27 AM. Reason: Fixed some latex
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    \int \frac{x}{\sin ^2x}dx=-arcotx -\int -arccot x dx

    u=x u'=1
    v'=\frac{1}{\sin ^2x}dx


    how to solve integral of arc cot??
    \int u \, dv = uv - \int v \, du.

    u = x \Rightarrow du = dx

    dv = \frac{1}{\sin^2 x} \, dx \Rightarrow v = - \cot x.

    So your integral becomes -x \cot x + \int \cot x \, dx, which is simple to find.
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    thanks
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