calculate-A-n-0-1-1-t-2-n-dt-with-n-integr-natural-

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 \dots

Answered by math khazana by abdo last updated on 09/Aug/18

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 \dots . . (2 n)}{1.3.5 \dots \dots (2 n - 1)} = \frac{2^{n} (n!) 2.4.8 \dots . . (2 n)}{1.2.3.4.5 \dots . . (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