Question

Fie sirul (In)n>0, In=integrala de la -1 la 1 din (1-x^2)^n dx.
Sa se demonstreze ca I(n+1)=(2n+2)In/2n+3.

Answer 1

$I_{n+1}=\int_{-1}^1(1-x^2)^{n+1}dx=\int_{-1}^1(1-x^2)^n(1-x^2)dx=$

$=\int_{-1}^1(1-x^2)^ndx-\int_{-1}^1(1-x^2)^nx^2dx$

Prima integrala este $I_n$, iar pe a doua o integrezi prin parti, luand:

$f(x)=x,\ f'(x)=1\ si\ \ g'(x)=(1-x^2)^nx\Rightarrow g(x)=-\dfrac12\cdot\dfrac{(1-x^2)^{n+1}}{n+1}$

Se obtine:

$I_{n+1}=I_n-\dfrac x2\cdot\dfrac{(1-x^2)^{n+1}}{n+1}\ |_{-1}^1-\dfrac{1}{2(n+1)}I_{n+1}$

Dupa inlocuire cu limitele de integrare se obtine:

$I_{n+1}(1+\dfrac{1}{2n+2})=I_n\Rightarrow I_{n+1}\cdot\dfrac{2n+3}{2n+2}=I_n$