show that ∫e^(sin(x)) dx= −Σ_(n=0) ^∞ (1/(n!))[ cos(x)∗(sin(x))^(n+1) ∗[(sin(x))^2 ]^((((−n)/2)−(1/2))) ∗ 2F_1 [(1/2),((1−n)/2);(3/2);(cos(x))^2 ] ]+c notice\2F_1 is special function called hypergeometric function


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Question Number 87793 by M±th+et£s last updated on 06/Apr/20
$show that ∫e^(sin(x)) dx= −Σ_(n=0) ^∞ (1/(n!))[ cos(x)∗(sin(x))^(n+1) ∗[(sin(x))^2 ]^((((−n)/2)−(1/2))) ∗ 2F_1 [(1/2),((1−n)/2);(3/2);(cos(x))^2 ] ]+c notice\2F_1 is special function called hypergeometric function$
$s h o w t h a t$ $\int e^{s i n (x)} d x =$ $- \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} [c o s (x) * {(s i n (x))}^{n + 1} * {[{(s i n (x))}^{2}]}^{(\frac{- n}{2} - \frac{1}{2})} * 2 F_{1} [\frac{1}{2}, \frac{1 - n}{2}; \frac{3}{2}; {(c o s (x))}^{2}]] + c$ $n o t i c e ∖ 2 F_{1} i s s p e c i a l f u n c t i o n c a l l e d h y p e r g e o m e t r i c f u n c t i o n$
	Answered by mind is power last updated on 06/Apr/20

	$\int {s i n}^{m} (x) d x$ $= \int \frac{u^{m}}{\sqrt{1 - u^{2}}} d u \int \sum_{n ⩾ 0} u^{m} . \frac{(2 n)! u^{2 n}}{2^{2 n} {(n!)}^{2}} d u = \sum_{n ⩾ 0} \int \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} u^{2 n + m} d u$ $= \sum_{n ⩾ 0} \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} . \frac{u^{2 n + m + 1}}{(2 n + m + 1)} + c$ $= u^{m + 1} (\frac{1}{m + 1} + \sum_{n ⩾ 1} \frac{2^{2 n} \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2})}{2 \underset{k = 0}{\prod^{n - 1}} (k + \frac{m + 3}{2})} \frac{u^{2 n}}{n!})$ $= u^{m + 1}$ $= \frac{u^{m + 1}}{m + 1} . \sum_{n ⩾ 0} \frac{\underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2}) . \underset{k = 0}{\prod^{n - 1}} (k + \frac{m + 1}{2})}{\underset{k = 0}{\prod^{n - 1}} (k + \frac{m + 3}{2})} . \frac{u^{2 n}}{n!}$ $= \frac{u^{m + 1}}{m + 1} . 2 F_{1} (\frac{1}{2}, \frac{m + 1}{2}; \frac{m + 3}{2}; u^{2})$ $e^{s i n (u)} = \sum_{m ⩾ 0} \frac{{s i n}^{m} (u)}{m!}$ $\int e^{s i n (u)} d u = \int \sum_{m ⩾ 0} \frac{{s i n}^{m} (u)}{m!} d u$ $= \sum_{m ⩾ 0} \frac{1}{m!} \int {s i n}^{m} (u) d u = \sum_{m ⩾ 0} \frac{{s i n}^{m + 1} (u)}{m! (m + 1)} . 2 F_{1} (\frac{1}{2}, \frac{m + 1}{2}; \frac{m + 3}{2}; {s i n}^{2} (u))$ $= \sum_{m ⩾ 0} \frac{{s i n}^{m + 1} (u)}{(m + 1)!} 2 F_{1} (\frac{1}{2}, \frac{m + 1}{2}; \frac{m + 3}{2}; {s i n}^{2} (u)) + c$ $m a y b e e m i s t a k s$
	Commented by M±th+et£s last updated on 06/Apr/20

	$m a y b e e i m a k e a m i s t a k e i w i l l p o s t m y$ $s o l u t i o n$
	Answered by M±th+et£s last updated on 06/Apr/20

	$e^{u} = \underset{n = 0}{\sum^{\infty}}, u = s i n (x) \Rightarrow\Rightarrow e^{s i n (x)} = \underset{n = 0}{\sum^{\infty}} \frac{{(s i n (x))}^{n}}{n!}$ $I = \int e^{s i n (x)} d x = \int \underset{n = 0}{\sum^{\infty}} \frac{{(s i n (x))}^{n}}{n!} d x = \underset{n = 0}{\sum^{\infty}} \int {(s i n (x))}^{n} d x$ $\int {(s i n (x))}^{n} d x = - c o s (x) * {(s i n (x))}^{n + 1} * {[s i n {(x)}^{2}]}^{\frac{- n}{2} - \frac{1}{2}} * 2 F_{1} [\frac{1}{2}, \frac{1 - n}{2}; \frac{3}{2}; {(c o s (x))}^{2}] o$ $\int e^{s i n (x)} d x = - \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} [c o s (x) * s i n {(x)}^{n + 1} * {[{(s i n (x))}^{2}]}^{\frac{- n}{2} - \frac{1}{2}} * 2 F_{1} [\frac{1}{2}, \frac{1 - n}{2}; \frac{3}{2}; (i c o s {(x)}^{2})]]$
	Commented by M±th+et£s last updated on 06/Apr/20

	$t h e r e i s a t y p o$ $i t s [{(c o s (x))}^{2}] n o t [(i c o s {(x)}^{2})]$
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