[tex]I_{n+1}=\int_{-1}^1(1-x^2)^{n+1}dx=\int_{-1}^1(1-x^2)^n(1-x^2)dx=[/tex]
[tex]=\int_{-1}^1(1-x^2)^ndx-\int_{-1}^1(1-x^2)^nx^2dx[/tex]
Prima integrala este [tex]I_n[/tex], iar pe a doua o integrezi prin parti, luand:
[tex]f(x)=x,\ f'(x)=1\ si\ \ g'(x)=(1-x^2)^nx\Rightarrow g(x)=-\dfrac12\cdot\dfrac{(1-x^2)^{n+1}}{n+1}[/tex]
Se obtine:
[tex]I_{n+1}=I_n-\dfrac x2\cdot\dfrac{(1-x^2)^{n+1}}{n+1}\ |_{-1}^1-\dfrac{1}{2(n+1)}I_{n+1}[/tex]
Dupa inlocuire cu limitele de integrare se obtine:
[tex]I_{n+1}(1+\dfrac{1}{2n+2})=I_n\Rightarrow I_{n+1}\cdot\dfrac{2n+3}{2n+2}=I_n[/tex]
