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Question Number 45885 by Meritguide1234 last updated on 17/Oct/18

Commented by maxmathsup by imad last updated on 18/Oct/18

$l e t A_{n} = \int_{0}^{\frac{π}{2}} n (1 - {(s i n x)}^{\frac{1}{n}}) d x = \int_{0}^{\frac{π}{2}} f_{n} (x) d x w i t h f_{n} (x) = n (1 - ({s i n x}^{\frac{1}{n}}))$ $b u t w e h a v e {(s i n x)}^{\frac{1}{n}} = e^{\frac{1}{n} l n (s i n x)} = 1 + \frac{l n (s i n x)}{n} + o (\frac{1}{n}) \Rightarrow$ $1 - e^{\frac{1}{n} l n (s i n x)} = - \frac{l n (s i n x)}{n} + o (\frac{1}{n}) \Rightarrow n (1 - e^{\frac{l n (s i n x)}{n}}) = - l n (s i n x) + o (1) \Rightarrow$ $f_{n} (x) = - l n (s i n x) + o (1) \Rightarrow {l i m}_{n \to + \infty} A_{n} = \int_{0}^{\frac{π}{2}} - l n (s i n x) d x$ $= - (- \frac{π}{2} l n (2)) = \frac{π}{2} l n (2) .$
Commented by Meritguide1234 last updated on 19/Oct/18

$n i c e a p p r o a c h$
Commented by maxmathsup by imad last updated on 20/Oct/18

$t h a n k y o u s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18
	$lim_(n→∞) n∫_0 ^(π/2) {1−(sinx)^(1/n) }dx lim_(n→∞) n[∣x∣_0 ^(π/2) −∫_0 ^(π/2) sin^(2p−1) xcoz^(2q−1) x dx] here 2p−1=(1/n) 2p=1+(1/n) p=((n+1)/(2n)) 2q−1=0 q=(1/2) using gamma beta function formula ∫_0 ^(π/2) sin^(2p−1) xcos^(2a−1) xdx=((⌈(p)⌈(q))/(2⌈(p+q))) lim_(n→∞) n[(π/2)−((⌈(p)⌈(q))/(2⌈(p+q)))] lim_(n→∞) n[(π/2)−((⌈(((n+1)/(2n)))⌈((1/2)))/(2⌈(((n+1)/(2n))+(1/2))))] lim_(n→∞) n[(π/2)−((⌈(1+((1−n)/(2n)))×(√π))/(2⌈(1+((n+1)/(2n))−(1/2)))) ] lim_(n→∞) n[(π/2)−(((√π) ×⌈(1+((1−n)/(2n))))/(2⌈(1+(1/(2n)))))]$
	$\lim_{n \to \infty} n \int_{0}^{\frac{π}{2}} {1 - {(s i n x)}^{\frac{1}{n}}} d x$ $\lim_{n \to \infty} n [∣ x ∣_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} {x c o z}^{2 q - 1} x d x]$ $h e r e 2 p - 1 = \frac{1}{n} 2 p = 1 + \frac{1}{n} p = \frac{n + 1}{2 n}$ $2 q - 1 = 0 q = \frac{1}{2} u s i n g g a m m a b e t a f u n c t i o n$ $f o r m u l a \int_{0}^{\frac{π}{2}} {s i n}^{2 p - 1} {x c o s}^{2 a - 1} x d x = \frac{⌈ (p) ⌈ (q)}{2 ⌈ (p + q)}$ $\lim_{n \to \infty} n [\frac{π}{2} - \frac{⌈ (p) ⌈ (q)}{2 ⌈ (p + q)}]$ $\lim_{n \to \infty} n [\frac{π}{2} - \frac{⌈ (\frac{n + 1}{2 n}) ⌈ (\frac{1}{2})}{2 ⌈ (\frac{n + 1}{2 n} + \frac{1}{2})}]$ $\lim_{n \to \infty} n [\frac{π}{2} - \frac{⌈ (1 + \frac{1 - n}{2 n}) \times \sqrt{π}}{2 ⌈ (1 + \frac{n + 1}{2 n} - \frac{1}{2})}]$ $\lim_{n \to \infty} n [\frac{π}{2} - \frac{\sqrt{π} \times ⌈ (1 + \frac{1 - n}{2 n})}{2 ⌈ (1 + \frac{1}{2 n})}]$
	Commented by Meritguide1234 last updated on 18/Oct/18

	$n o t c o r r e c t$
	Commented by tanmay.chaudhury50@gmail.com last updated on 18/Oct/18

	$y e s i k n o w t h e r e s u l t i s y e t t o c o m e . . . w h e n n \to \infty$ $t h e r e s u l t s i s \infty$
	Commented by Meritguide1234 last updated on 18/Oct/18

	$p u t n = 1 / t . . . t h e n a n s i s (\frac{π}{2} l n 2)$
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