a) Tính $ I = \int\limits_{0}^{1}x(1-x^2)^ndx, n \in  N$
b) Chứng minh rằng $\frac{1}{2} C^0_n - \frac{1}{4} C^1_n + \frac{1}{6}C^2_n-...+\frac{(-1)^n}{2n+2}C^n_n = \frac{1}{2n+2}, \forall n \in  N$  
a) Đặt $t = 1 - x^2    \Rightarrow   dt = - 2xdx   \Rightarrow   xdx = -\frac{dt}{2} $

Đổi cận:


$I = \int\limits_{1}^{0}t^n \left ( -\frac{dt}{2}   \right ) = \frac{1}{2} \int\limits_{0}^{1}t^ndt = \frac{1}{2} \left ( \frac{t^{n+1}}{n+1}  \right )\left| \begin{array}{l}
1\\
0
\end{array} \right. = \frac{1}{2(n+1)}.$

b) Ta có : $x(1-x^2)^n = xC^0_n - x^3C^1_n + x^5C^2_n - ... + (-1)^nx^{2n+1}C^n_n, \forall x \in  R$
$\Rightarrow \int\limits_{0}^{1}x(1-x^2)^ndx = \left ( \frac{x^2}{2}C^0_n -\frac{x^4}{4}C^1_n + \frac{x^6}{6}C^2_n-...+(-1)^n\frac{x^{2n+2}}{2n+2}C^n_n  \right )\left| \begin{array}{l}
1\\
0
\end{array} \right.$
$\Rightarrow \frac{1}{2n+2} = \frac{1}{2} C^0_n - \frac{1}{4}C^1_n + \frac{1}{6}C^2_n-...+\frac{(-1)^n}{2n+2}C^n_n.$