integration question no. 9

**afeasfaerw23231233** · November 21st 2007, 12:38 AM

**Krizalid** · November 21st 2007, 04:05 AM

Originally Posted by afeasfaerw23231233

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.

**afeasfaerw23231233** · November 22nd 2007, 01:14 AM

thanks

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