# Thread: integration question no. 9

1. ## integration question no. 9

integration question no. 9

2. Originally Posted by afeasfaerw23231233
This just needs a little trick to finish it:

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

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

Now

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

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

Back to the problem

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

Multiply and we're done.

3. thanks