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Math Help - integration question no. 9

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    integration question no. 9

    integration question no. 9
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    Quote Originally Posted by afeasfaerw23231233 View Post
    This just needs a little trick to finish it:

    \left( {1 - x^2 } \right)^n - 2nx^2 \left( {1 - x^2 } \right)^{n - 1} = \left( {1 - x^2 } \right)^{n - 1} \left[ {\left( {1 - x^2 } \right) - 2nx^2 } \right].

    Let's focus on {\left( {1 - x^2 } \right) - 2nx^2 }.

    Now

    \left( {1 - x^2 } \right) - 2nx^2 = \left( {2n - 2nx^2 } \right) + \left( {1 - x^2 } \right) - 2n = 2n\left( {1 - x^2 } \right) + \left( {1 - x^2 } \right) - 2n.

    So \left( {1 - x^2 } \right) - 2nx^2 = \left( {1 - x^2 } \right)\left( {2n + 1} \right) - 2n.

    Back to the problem

    f'(x) = \left( {1 - x^2 } \right)^{n - 1} \left[ {\left( {1 - x^2 } \right)\left( {2n + 1} \right) - 2n} \right].

    Multiply and we're done.
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    thanks
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