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Question Number 48718 by Abdo msup. last updated on 27/Nov/18

l e t I_{n} = \int_{0}^{1} {(1 - t^{2})}^{n} d t

1) c a l c u l a t e I_{n} b y r e c u r r e n c e

2) f i n d t h e v a l u e o f \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 1} C_{n}^{k}

Commented by Abdo msup. last updated on 27/Nov/18

w e h a v e I_{n} = \int_{0}^{1} {(1 - t^{2})}^{n - 1} (1 - t^{2}) d t

= \int_{0}^{1} {(1 - t^{2})}^{n - 1} - \int_{0}^{1} t^{2} {(1 - t^{2})}^{n - 1} d t

b u t \int_{0}^{1} {(1 - t^{2})}^{n - 1} d t = I_{n - 1}

b y p a r t s \int_{0}^{1} t^{2} {(1 - t^{2})}^{n - 1} d t

= - \frac{1}{2} \int_{0}^{1} t (- 2 t) {(1 - t^{2})}^{n - 1} d t u = t a n d v^{^{'}} = (- 2 t) {(1 - t^{2})}^{n - 1}

= - \frac{1}{2} {{[\frac{1}{n} {(1 - t^{2})}^{n} t]}_{0}^{1} - \int_{0}^{1} \frac{1}{n} {(1 - t^{2})}^{n} d t}

= \frac{1}{2 n} I_{n} \Rightarrow I_{n} = I_{n - 1} - \frac{1}{2 n} I_{n} \Rightarrow

(1 + \frac{1}{2 n}) I_{n} = I_{n - 1} \Rightarrow \frac{2 n + 1}{2 n} I_{n} = I_{n - 1} \Rightarrow I_{n} = \frac{2 n}{2 n + 1} I_{n - 1} \Rightarrow

\prod_{k = 1}^{n} I_{k} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 1} \prod_{k = 1}^{n} I_{k - 1} \Rightarrow

I_{n} = \prod_{k = 1}^{n} \frac{2 k}{2 k + 1} I_{o} b u t I_{0} = 1 \Rightarrow I_{n} = \frac{\prod_{k = 1}^{n} (2 k)}{\prod_{k = 1}^{n} (2 k + 1)}

= \frac{2^{n} n!}{3.5.7 \dots . . (2 n + 1)} = \frac{2^{n} n!}{2.3.4.5 \dots (2 n) (2 n + 1)} \times 2^{n} n!

= \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!}

Commented by Abdo msup. last updated on 27/Nov/18

2) w e h a v e I_{n} = \int_{0}^{1} {(1 - t^{2})}^{n} d t

= \int_{0}^{1} \sum_{k = 0}^{n} C_{n}^{k} {(- t^{2})}^{k} d t = \sum_{k = 0}^{n} {(- 1)}^{k} C_{n}^{k} \int_{0}^{1} t^{2 k} d t

= \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 1} C_{n}^{k} \Rightarrow \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2 k + 1} C_{n}^{k} = \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!} .

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18

\int_{0}^{1} {(1 - t^{2})}^{n} d t

t = s i n θ d t = c o s θ d θ

\int_{0}^{\frac{π}{2}} {c o s}^{2 n} θ \times c o s θ d θ

\int_{0}^{\frac{π}{2}} {(s i n θ)}^{2 \times \frac{1}{2} - 1} \times {(c o s θ)}^{2 n + 2 - 1} d θ

u s i n g g a m m a b e t a f u n c t i o n

= \frac{⌈ (\frac{1}{2}) \times ⌈ (2 n + 2)}{2 ⌈ (2 n + 2 + \frac{1}{2})}

= \frac{\sqrt{π} \times ⌈ (2 n + 2)}{2 ⌈ (2 n + \frac{5}{2})}

f o r m u l a 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} θ {c o s}^{2 q - 1} θ d θ

= \frac{⌈ (p) ⌈ (q)}{⌈ (p + q)}