calculate A_n = ∫_0 ^1 (1−t^2 )^n dt with n integr natural


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Question Number 41518 by maxmathsup by imad last updated on 08/Aug/18

$c a l c u l a t e A_{n} = \int_{0}^{1} {(1 - t^{2})}^{n} d t w i t h n i n t e g r n a t u r a l$
	Answered by alex041103 last updated on 09/Aug/18

	$W e k n o w t h a t$ ${(a + b)}^{n} = \underset{k = 0}{\sum^{n}} (\begin{matrix} k \\ n \end{matrix}) a^{n - k} b^{k}$ $\Rightarrow {(1 - t^{2})}^{n} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k} (\begin{matrix} k \\ n \end{matrix}) t^{2 k}$ $\Rightarrow A_{n} = \underset{0}{\int^{1}} [\underset{k = 0}{\sum^{n}} {(- 1)}^{k} (\begin{matrix} k \\ n \end{matrix}) t^{2 k}] d t =$ $= \underset{k = 0}{\sum^{n}} [{(- 1)}^{k} (\begin{matrix} k \\ n \end{matrix}) \underset{0}{\int^{1}} t^{2 k} d t] =$ $= \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{2 k + 1} (\begin{matrix} k \\ n \end{matrix})$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$e x c e l l e n t . . .$
	Answered by math khazana by abdo last updated on 09/Aug/18
	$we have A_(n+1) =∫_0 ^1 (1−t^2 )^(n+1) dt =∫_0 ^1 (1−t^2 )^n (1−t^2 )dt= ∫_0 ^1 (1−t^2 )^n −∫_0 ^1 t^2 (1−t^2 )^n dt =A_n −∫_0 ^1 t^2 (1−t^2 )^n dt by parts ∫_0 ^1 t^2 (1−t^2 )^n dt =−(1/2) ∫_0 ^1 t(−2t)(1−t^2 )^n dt =−(1/2){ [ (t/(n+1))(1−t^2 )^(n+1) ]_0 ^1 −∫_0 ^1 (1/(n+1))(1−t^2 )^(n+1) dt} =(1/(2(n+1))) A_(n+1) ⇒(1−(1/(2(n+1))))A_(n+1) =A_n ⇒ ((2n+1)/(2n+2)) A_(n+1) = A_n ⇒A_(n+1) =((2n+2)/(2n+1)) A_n ⇒ Π_(k=0) ^(n−1) A_(k+1) =Π_(k=0) ^(n−1) ((2k+2)/(2k+1)) Π_(k=0) ^(n−1) A_n ⇒ A_n = Π_(k=0) ^(n−1) ((2k+2)/(2k+1)) A_(0 ) but A_0 =1 ⇒ A_n =((2.4.8.....(2n))/(1.3.5......(2n−1))) =((2^n (n!) 2.4.8.....(2n))/(1.2.3.4.5.....(2n))) =((2^(2n) (n!)^2 )/((2n)!)) .$
	$w e h a v e A_{n + 1} = \int_{0}^{1} {(1 - t^{2})}^{n + 1} d t$ $= \int_{0}^{1} {(1 - t^{2})}^{n} (1 - t^{2}) d t = \int_{0}^{1} {(1 - t^{2})}^{n} - \int_{0}^{1} t^{2} {(1 - t^{2})}^{n} d t$ $= A_{n} - \int_{0}^{1} t^{2} {(1 - t^{2})}^{n} d t b y p a r t s$ $\int_{0}^{1} t^{2} {(1 - t^{2})}^{n} d t = - \frac{1}{2} \int_{0}^{1} t (- 2 t) {(1 - t^{2})}^{n} d t$ $= - \frac{1}{2} {{[\frac{t}{n + 1} {(1 - t^{2})}^{n + 1}]}_{0}^{1} - \int_{0}^{1} \frac{1}{n + 1} {(1 - t^{2})}^{n + 1} d t}$ $= \frac{1}{2 (n + 1)} A_{n + 1} \Rightarrow (1 - \frac{1}{2 (n + 1)}) A_{n + 1} = A_{n} \Rightarrow$ $\frac{2 n + 1}{2 n + 2} A_{n + 1} = A_{n} \Rightarrow A_{n + 1} = \frac{2 n + 2}{2 n + 1} A_{n} \Rightarrow$ $\prod_{k = 0}^{n - 1} A_{k + 1} = \prod_{k = 0}^{n - 1} \frac{2 k + 2}{2 k + 1} \prod_{k = 0}^{n - 1} A_{n} \Rightarrow$ $A_{n} = \prod_{k = 0}^{n - 1} \frac{2 k + 2}{2 k + 1} A_{0} b u t A_{0} = 1 \Rightarrow$ $A_{n} = \frac{2.4.8 . . . . . (2 n)}{1.3.5 . . . . . . (2 n - 1)} = \frac{2^{n} (n!) 2.4.8 . . . . . (2 n)}{1.2.3.4.5 . . . . . (2 n)}$ $= \frac{2^{2 n} {(n!)}^{2}}{(2 n)!} .$
	Commented by alex041103 last updated on 10/Aug/18

	$P e r f e c t! D e s e r v e s a l i k e!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$l e t t = s i n \propto d t = c o s \propto d \propto$ $\int_{0}^{\frac{Π}{2}} {c o s}^{2 n} \propto c o s \propto d \propto$ $\int_{0}^{\frac{Π}{2}} {c o s}^{2 n + 1} \propto . {s i n}^{0} \propto d \propto$ $\int_{0}^{\frac{Π}{2}} {c o s}^{2 (n + 1) - 1} . {s i n}^{2 . \frac{1}{2} - 1} \propto d \propto$ $\frac{⌈ (2 n + 2) ⌈ (\frac{1}{2})}{2 ⌈ (2 n + 2 + \frac{1}{2})} ⌈ (n) = g a m m a f u n c t i o n$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

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