I-n-0-1-x-n-x-2-1-dx-find-reduction-formula-

Question Number 138209 by 676597498 last updated on 11/Apr/21

I_{n} = \int_{0}^{1} x^{n} \sqrt{x^{2} + 1} d x

f i n d r e d u c t i o n f o r m u l a

Commented by Dwaipayan Shikari last updated on 11/Apr/21

\sqrt{x^{2} + 1} = \underset{k = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{k}}{k!} {(- x^{2})}^{k}

= \int_{0}^{1} x^{n} \sqrt{x^{2} + 1} d x = \underset{k = 0}{\sum^{\infty}} \int_{0}^{1} \frac{{(- \frac{1}{2})}_{k}}{k!} {(- 1)}^{k} x^{2 k + n} d x

= \underset{k = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{k}}{k! (2 k + n + 1)} {(- 1)}^{k} = \frac{1}{2} \underset{k = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{k}}{k! (k + \frac{n + 1}{2})} {(- 1)}^{k}

= \frac{1}{2} \underset{k = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{k} Γ (k + \frac{n + 1}{2})}{k! Γ (k + \frac{n + 3}{2})} {(- 1)}^{k} = \frac{Γ (\frac{n + 1}{2})}{2 Γ (\frac{n + 3}{2})} \underset{k = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{k} {(\frac{n + 1}{2})}_{k}}{{(\frac{n + 3}{2})}_{k}} {(- 1)}^{k}

= \frac{1}{n + 1}_{2} F_{1} (- \frac{1}{2}, \frac{n + 1}{2}; \frac{n + 3}{2}; - 1)

E x a m p l e w h e n n = 0

_{2} F_{1} (- \frac{1}{2}, \frac{1}{2}; \frac{3}{2}; - 1) = \frac{Γ (\frac{3}{2})}{Γ (\frac{1}{2}) Γ (1)} \int_{0}^{1} x^{- \frac{1}{2}} {(1 - x)}^{\frac{3}{2} - \frac{1}{2} - 1} {(1 + x)}^{\frac{1}{2}} d x

= \frac{1}{2} \int_{0}^{1} x^{- \frac{1}{2}} {(1 + x)}^{\frac{1}{2}} d x = \int_{0}^{1} \sqrt{1 + x^{2}} d x = \frac{2 \sqrt{2}}{3}

Answered by physicstutes last updated on 11/Apr/21

I_{n} = \int_{0}^{1} x^{n} \sqrt{x^{2} + 1} d x

I_{n} = \int_{0}^{1} x^{n - 1} x \sqrt{x^{2} + 1} d x

let u = x^{n - 1} and d v = x \sqrt{x^{2} + 1} d x

\Rightarrow d u = (n - 1) x^{n - 2} d x and v = \frac{1}{2} (\frac{2}{3}) {(x^{2} + 1)}^{3 / 2}

∴ I_{n} = {[\frac{x^{n - 1}}{3} {(x^{2} + 1)}^{\frac{3}{2}}]}_{0}^{1} - \frac{(n - 1)}{3} \int_{0}^{1} x^{n - 2} (x^{2} + 1) \sqrt{x^{2} + 1} d x

I_{n} = \frac{2^{3 / 2}}{3} - \frac{(n - 1)}{3} [\int_{0}^{1} x^{n} \sqrt{x^{2} + 1} d x + \int_{0}^{1} x^{n - 2} \sqrt{x^{2} + 1} d x]

I_{n} = \frac{2 \sqrt{2}}{3} - \frac{(n - 1)}{3} I_{n} - \frac{(n - 1)}{3} I_{n - 2}

\Rightarrow I_{n} (\frac{n + 2}{3}) = \frac{2 \sqrt{2}}{3} - \frac{(n - 1)}{3} I_{n - 2}

hence I_{n} = \frac{2 \sqrt{2} - (n - 1) I_{n - 2}}{(n + 2)}, n ⩾ 2

Answered by mathmax by abdo last updated on 12/Apr/21

I_{n} = \int_{0}^{1} x^{n} \sqrt{1 + x^{2}} dx changement x = sht give

I_{n} = \int_{0}^{argsh (1)} {sh}^{n} t ch (t) ch (t) dt

= \int_{0}^{\ln (1 + \sqrt{2})} {sh}^{n} t cht (cht) dt (u^{^{'}} = cht {sh}^{n} t and v = cht)

= {[\frac{1}{n + 1} {sh}^{n + 1} t cht]}_{0}^{\ln (1 + \sqrt{2})} - \frac{1}{n + 1} \int_{0}^{\ln (1 + \sqrt{2})} {sh}^{n + 2} t dt

= \frac{1}{n + 1} {{\frac{e^{t} + e^{- t}}{2} {(\frac{e^{t} - e^{- t}}{2})}^{n + 1}]}_{0}^{\ln (1 + \sqrt{2})} - J_{n}

= \frac{1}{n + 1) 2^{n + 2}} {(1 + \sqrt{2} + {(1 + \sqrt{2})}^{- 1}) {(1 + \sqrt{2} - {(1 + \sqrt{2})}^{- 1})}^{n + 1}} - J_{n}

(n + 1) J_{n} = \int_{0}^{\ln (1 + \sqrt{2})} {(\frac{e^{t} - e^{- t}}{2})}^{n + 2} dt

= \frac{1}{2^{n + 2}} \int_{0}^{\ln (1 + \sqrt{2})} \sum_{k = 0}^{n + 2} C_{k}^{n + 2} {(e^{t})}^{k} {(- e^{- t})}^{n + 2 - k} dt

= \frac{1}{2^{n + 2}} \sum_{k = 0}^{n + 2} C_{n + 2}^{k} {(- 1)}^{n + 2 - k} \int_{0}^{\ln (1 + \sqrt{2})} e^{kt} . e^{- (n + 2 - k) t} dt

= \frac{1}{2^{n + 2}} \sum_{k = 0}^{n + 2} C_{n + 2}^{k} {(- 1)}^{n - k} \int_{0}^{\ln (1 + \sqrt{2})} e^{(2 k - n - 2) t} dt

and \int_{0}^{\ln (1 + \sqrt{2})} e^{(2 k - n - 2) t} = {[\frac{1}{2 k - n - 2} e^{(2 k - n - 2) t}]}_{0}^{\ln (1 + \sqrt{2})}

= \frac{1}{2 k - n - 2} ({(1 + \sqrt{2})}^{2 k - n - 2} - 1) \Rightarrow

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

so the value of I_{n} is known \dots