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](https://www.tinkutara.com/question/Q87793.png)
Answered by mind is power last updated on 06/Apr/20

Commented by M±th+et£s last updated on 06/Apr/20

Answered by M±th+et£s last updated on 06/Apr/20
![e^u =Σ_(n=0) ^∞ , u=sin(x)⇒⇒e^(sin(x)) =Σ_(n=0) ^∞ (((sin(x))^n )/(n!)) I=∫e^(sin(x)) dx=∫Σ_(n=0) ^∞ (((sin(x))^n )/(n!))dx=Σ_(n=0) ^∞ ∫(sin(x))^n dx ∫(sin(x))^n dx=−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 ]o ∫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);(icos(x)^2 )]]](https://www.tinkutara.com/question/Q87846.png)
Commented by M±th+et£s last updated on 06/Apr/20
![there is a typo its [(cos(x))^2 ] not [(icos(x)^2 )]](https://www.tinkutara.com/question/Q87848.png)