Question

Se considera sirul (I indice n )n$\geq$ 0 definit astfel:
I indice 0= $\int\limits^\frac{\pi }{2} _0 {sinx} \, dx$ si I indice n= $\int\limits^\frac{\pi }{2} _0 {x^n sinx} \, dx$, pentru orice n$\geq$ 1.

Demonstrati ca I indice n= n[ $\frac{\pi }{2}$ la puterea n-1] -n(n-1)I indice n-2, oricare ar fi n$\geq$ 2.

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Answer 1

$\displaystyle (I_n)_{n\geq 0}, I_n = \int_{0}^{\frac{\pi}{2}}{x^n\sin{x}dx}\\I_n = \int{x^n\sin{x}dx}\\\\\textrm{Integrare prin parti:}\\\\u = x^n\implies du = nx^{n-1}dx\\\\dv = \sin{x}dx \implies v' = \sin{x}\implies v = -\cos{x}\\\\I_n = uv - \int{v\:du} = x^n\cdot (-\cos{x}) - n\int{-\cos{x}\cdot x^{n-1}dx} = n\int{\cos{x}\cdot x^{n-1}dx} - x^n\cdot \cos{x}$

$\displaystyle \textrm{Acum integram si }\int{\cos{x}\cdot x^{n-1}dx}\textrm{ prin parti.}\\u_2 = x^{n-1}\implies du_2 = (n-1)x^{n-2}dx\\dv_2 = \cos{x}dx\implies v_2' = \cos{x} \implies v_2 = \sin{x}$

$\displaystyle \int{\cos{x}\cdot x^{n-1}dx} = u_2v_2 - \int{v_2\:du_2} = x^{n-1}\sin{x} - \int{\sin{x}\cdot (n-1)x^{n-2}dx} = x^{n-1}\sin{x} - (n-1)\int{\sin{x}\cdot x^{n-2}dx} = x^{n-1}\sin{x} - (n-1)\cdot I_{n-2}$

$\displaystyle I_n = n\int{\cos{x}\cdot x^{n-1}dx} - x^n\cdot \cos{x} \\\\= n\cdot (x^{n-1}\sin{x} - (n-1)\cdot I_{n-2}) - x^n\cdot \cos{x} \\\\= n\cdot x^{n-1}\sin{x} - n(n-1)\cdot I_{n-2} - x^n\cdot \cos{x} \Big {|_{0}^{\frac{\pi}{2}}}\\\\ = n\cdot (\frac{\pi}{2})^{n-1}\sin{\frac{\pi}{2}} - (\frac{\pi}{2})^n\cdot \cos{\frac{\pi}{2}} - (n\cdot 0^{n-1}\sin{0} - 0^n\cdot \cos{0}) - n(n-1)\cdot I_{n-2} \\\\I_n= n\cdot (\frac{\pi}{2})^{n-1}\cdot 1 - (\frac{\pi}{2})^n\cdot 0 - (n\cdot 0^{n-1}\cdot 0 - 0^n\cdot 1) - n(n-1)\cdot I_{n-2} \\\\\boxed{I_n= n\cdot (\frac{\pi}{2})^{n-1} - n(n-1)\cdot I_{n-2}}$