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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
showthatesin(x)dx=n=01n![cos(x)(sin(x))n+1[(sin(x))2](n212)2F1[12,1n2;32;(cos(x))2]]+cnotice2F1isspecialfunctioncalledhypergeometricfunction
Answered by mind is power last updated on 06/Apr/20
∫sin^m (x)dx  =∫(u^m /( (√(1−u^2 ))))du∫Σ_(n≥0) u^m .(((2n)!u^(2n) )/(2^(2n) (n!)^2 ))du=Σ_(n≥0) ∫(((2n)!)/(2^(2n) (n!)^2 ))u^(2n+m) du  =Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 )).(u^(2n+m+1) /((2n+m+1)))+c  =u^(m+1) ((1/(m+1))+Σ_(n≥1) ((2^(2n) Π_(k=0) ^(n−1) (k+(1/2)))/(2Π_(k=0) ^(n−1) (k+((m+3)/2))))(u^(2n) /(n!)))  =u^(m+1)   =(u^(m+1) /(m+1)).Σ_(n≥0) ((Π_(k=0) ^(n−1) (k+(1/2)).Π_(k=0) ^(n−1) (k+((m+1)/2)))/(Π_(k=0) ^(n−1) (k+((m+3)/2)))).(u^(2n) /(n!))  =(u^(m+1) /(m+1)).2F_1 ((1/2),((m+1)/2);((m+3)/2);u^2 )  e^(sin(u)) =Σ_(m≥0) ((sin^m (u))/(m!))  ∫e^(sin(u)) du=∫Σ_(m≥0) ((sin^m (u))/(m!))du  =Σ_(m≥0) (1/(m!))∫sin^m (u)du=Σ_(m≥0) ((sin^(m+1) (u))/(m!(m+1))).2F_1 ((1/2),((m+1)/2);((m+3)/2);sin^2 (u))  =Σ_(m≥0) ((sin^(m+1) (u))/((m+1)!))2F_1 ((1/2),((m+1)/2);((m+3)/2);sin^2 (u))+c  may bee mistaks
sinm(x)dx=um1u2dun0um.(2n)!u2n22n(n!)2du=n0(2n)!22n(n!)2u2n+mdu=n0(2n)!22n(n!)2.u2n+m+1(2n+m+1)+c=um+1(1m+1+n122nn1k=0(k+12)2n1k=0(k+m+32)u2nn!)=um+1=um+1m+1.n0n1k=0(k+12).n1k=0(k+m+12)n1k=0(k+m+32).u2nn!=um+1m+1.2F1(12,m+12;m+32;u2)esin(u)=m0sinm(u)m!esin(u)du=m0sinm(u)m!du=m01m!sinm(u)du=m0sinm+1(u)m!(m+1).2F1(12,m+12;m+32;sin2(u))=m0sinm+1(u)(m+1)!2F1(12,m+12;m+32;sin2(u))+cmaybeemistaks
Commented by M±th+et£s last updated on 06/Apr/20
may bee i make a mistake i will post my  solution
maybeeimakeamistakeiwillpostmysolution
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 )]]
eu=n=0,u=sin(x)⇒⇒esin(x)=n=0(sin(x))nn!I=esin(x)dx=n=0(sin(x))nn!dx=n=0(sin(x))ndx(sin(x))ndx=cos(x)(sin(x))n+1[sin(x)2]n2122F1[12,1n2;32;(cos(x))2]oesin(x)dx=n=01n![cos(x)sin(x)n+1[(sin(x))2]n2122F1[12,1n2;32;(icos(x)2)]]
Commented by M±th+et£s last updated on 06/Apr/20
there is a typo  its [(cos(x))^2 ] not [(icos(x)^2 )]
thereisatypoits[(cos(x))2]not[(icos(x)2)]

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