find I_n = ∫_0 ^1 x^n (√(1−x^2 ))dx


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Question Number 42798 by maxmathsup by imad last updated on 02/Sep/18

$f i n d I_{n} = \int_{0}^{1} x^{n} \sqrt{1 - x^{2}} d x$
Commented by maxmathsup by imad last updated on 05/Sep/18
$changement x=sint give I_n = ∫_0 ^(π/2) sin^n t cos^2 t dt =∫_0 ^(π/2) sin^n t(1−sin^2 t)dt = ∫_0 ^(π/2) sin^n t dt −∫_0 ^(π/2) sin^(n+2) dt =w_n −w_(n+2) let find w_n = ∫_0 ^(π/2) sin^n t dt we have w_n =∫_0 ^(π/2) sin^(n−2) t(1−cos^2 t)dt =w_(n−2) −∫_0 ^(π/2) cost( cost sin^(n−2) t)dt by parts u=cost and v^′ =cost sin^(n−2) t⇒ ∫_0 ^(π/2) cost (cost sin^(n−2) t)dt =[(1/(n−1)) cost sin^(n−1) t]_0 ^(π/2) −∫_0 ^(π/2) ((−sint)/(n−1)) sin^(n−1) tdt = ∫_0 ^(π/2) ((sin^n t)/(n−1)) dt =(1/(n−1)) w_n ⇒w_n =w_(n−2) −(1/(n−1)) w_n ⇒(1+(1/(n−1)))w_n =w_(n−2) ⇒ (n/(n−1))w_n =w_(n−2) ⇒w_n =((n−1)/n)w_(n−2) ⇒w_(2n) =((2n−1)/(2n))w_(2n−2) and w_(2n+1) =((2n)/(2n+1)) w_(2n−1) let find w_(2n) Π_(k=1) ^n w_(2k) =Π_(k=1) ^n ((2k−1)/(2k)) w_(2k−2) ⇒w_2 .w_4 .....w_(2n) =((1.3.5.....(2n−1))/((2)(4)....(2n)))w_0 w_2 ...w_(2n−2) ⇒ w_(2n) = ((1.3.5...(2n−1))/(2^n (n!))) w_0 = (((2n)!)/(2^(2n) (n!)^2 )) (π/2) let find w_(2n+1) I_(2n) =w_(2n) −w_(2n+2) = (((2n)!)/(2^(2n) (n!)^2 )) (π/2) − (((2n+4)!)/(2^(2n+4) {(n+2)!}^2 )) (π/2) let find w_(2n+1) we have w_(2n+1) =((2n)/(2n+1)) w_(2n−1) ⇒Π_(k=1) ^n w_(2k+1) =Π_(k=1) ^n ((2k)/(2k+1)) Π_(k=1) ^n w_(2k−1) ⇒ w_3 .w_5 .....w_(2n+1) =((2^n (n!))/(3.5....(2n+1))) w_1 .w_3 ....w_(2n−1) ⇒ w_(2n+1) = ((2^(2n) (n!)^2 )/((2n+1)!)) w_1 but w_1 =1 ⇒ I_(2n−1) =w_(2n−1) −w_(2n+1) =((2^(2n−2) {(n−1)!}^2 )/((2n−1)!)) −((2^(2n) (n!)^2 )/((2n+1)!)) .$
$c h a n g e m e n t x = s i n t g i v e I_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{n} t {c o s}^{2} t d t$ $= \int_{0}^{\frac{π}{2}} {s i n}^{n} t (1 - {s i n}^{2} t) d t = \int_{0}^{\frac{π}{2}} {s i n}^{n} t d t - \int_{0}^{\frac{π}{2}} {s i n}^{n + 2} d t = w_{n} - w_{n + 2}$ $l e t f i n d w_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{n} t d t w e h a v e w_{n} = \int_{0}^{\frac{π}{2}} {s i n}^{n - 2} t (1 - {c o s}^{2} t) d t$ $= w_{n - 2} - \int_{0}^{\frac{π}{2}} c o s t (c o s t {s i n}^{n - 2} t) d t b y p a r t s u = c o s t a n d v^{^{'}} = c o s t {s i n}^{n - 2} t \Rightarrow$ $\int_{0}^{\frac{π}{2}} c o s t (c o s t {s i n}^{n - 2} t) d t = {[\frac{1}{n - 1} c o s t {s i n}^{n - 1} t]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{- s i n t}{n - 1} {s i n}^{n - 1} t d t$ $= \int_{0}^{\frac{π}{2}} \frac{{s i n}^{n} t}{n - 1} d t = \frac{1}{n - 1} w_{n} \Rightarrow w_{n} = w_{n - 2} - \frac{1}{n - 1} w_{n} \Rightarrow (1 + \frac{1}{n - 1}) w_{n} = w_{n - 2} \Rightarrow$ $\frac{n}{n - 1} w_{n} = w_{n - 2} \Rightarrow w_{n} = \frac{n - 1}{n} w_{n - 2} \Rightarrow w_{2 n} = \frac{2 n - 1}{2 n} w_{2 n - 2} a n d$ $w_{2 n + 1} = \frac{2 n}{2 n + 1} w_{2 n - 1} l e t f i n d w_{2 n}$ $\prod_{k = 1}^{n} w_{2 k} = \prod_{k = 1}^{n} \frac{2 k - 1}{2 k} w_{2 k - 2} \Rightarrow w_{2} . w_{4} . . . . . w_{2 n} = \frac{1.3.5 . . . . . (2 n - 1)}{(2) (4) . . . . (2 n)} w_{0} w_{2} . . . w_{2 n - 2} \Rightarrow$ $w_{2 n} = \frac{1.3.5 . . . (2 n - 1)}{2^{n} (n!)} w_{0} = \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \frac{π}{2} l e t f i n d w_{2 n + 1}$ $I_{2 n} = w_{2 n} - w_{2 n + 2} = \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} \frac{π}{2} - \frac{(2 n + 4)!}{2^{2 n + 4} {(n + 2)!}^{2}} \frac{π}{2} l e t f i n d w_{2 n + 1}$ $w e h a v e w_{2 n + 1} = \frac{2 n}{2 n + 1} w_{2 n - 1} \Rightarrow \prod_{k = 1}^{n} w_{2 k + 1} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 1} \prod_{k = 1}^{n} w_{2 k - 1} \Rightarrow$ $w_{3} . w_{5} . . . . . w_{2 n + 1} = \frac{2^{n} (n!)}{3.5 . . . . (2 n + 1)} w_{1} . w_{3} . . . . w_{2 n - 1} \Rightarrow$ $w_{2 n + 1} = \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!} w_{1} b u t w_{1} = 1 \Rightarrow I_{2 n - 1} = w_{2 n - 1} - w_{2 n + 1}$ $= \frac{2^{2 n - 2} {(n - 1)!}^{2}}{(2 n - 1)!} - \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!} .$
Commented by maxmathsup by imad last updated on 05/Sep/18

$a n o t h e r w a y c h a n g e m e n t \sqrt{1 - x^{2}} = t g i v e x^{2} = 1 - t^{2} \Rightarrow d x = - d t \Rightarrow$ $I_{n} = \int_{0}^{1} {(1 - t^{2})}^{\frac{n}{2}} d t \Rightarrow I_{2 n} = \int_{0}^{1} {(1 - t^{2})}^{n} d t = \int_{0}^{1} \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} t^{2 k} d t$ $= \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} \frac{1}{2 k + 1} = \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 1} C_{n}^{k}$ $I_{2 n + 1} = \int_{0}^{1} {(1 - t^{2})}^{n + \frac{1}{2}} d t = \int_{0}^{1} {(1 - t^{2})}^{n} \sqrt{1 - t^{2}} d t$ $= \int_{0}^{1} \sqrt{1 - t^{2}} (\sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} t^{2 k}) d t$ $= \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} \int_{0}^{1} t^{2 k} \sqrt{1 - t^{2}} d t t h e n w e c a l c u l a t e w_{k} = \int_{0}^{1} t^{2 k} \sqrt{1 - t^{2}} d t$ $b y r e c u r r e n c e . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18

	$x = s i n α d x = c o s α d α$ $\int_{0}^{\frac{Π}{2}} {s i n}^{n} α {c o s}^{2} α d α$ $u s i n g g a m m a b e t a f u n c t i o n$ $\int_{0}^{\frac{Π}{2}} {s i n}^{2 p - 1} α {c o s}^{2 q - 1} \propto d α$ $= \frac{⌈ p ⌈ q}{2 ⌈ (p + q)}$ $2 p - 1 = n p = \frac{n + 1}{2} 2 q - 1 = 2 q = \frac{3}{2}$ $\int_{0}^{\frac{Π}{2}} {s i n}^{2 \times \frac{n + 1}{2} - 1} α \times {c o s}^{2 \times \frac{3}{2} - 1} α d α$ $= \frac{⌈ (\frac{n + 1}{2}) \times ⌈ (\frac{3}{2})}{2 ⌈ (\frac{n + 4}{2})}$ $p l s c h e c k$
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