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Question Number 86671 by M±th+et£s last updated on 30/Mar/20

∫sin(x) arcsin(x)

sin(x)arcsin(x)

Answered by Rio Michael last updated on 30/Mar/20

let me give a try.   ∫ sin (x) arcsin(x) dx  using the taylor series expansion for sin (x) arcsin (x) centred at 0   sin (x) arcsin (x) = x^2  + (x^6 /(18)) + (x^8 /(30)) + ((2669 x^(10) )/(113400)) + ((601 x^(12) )/(34020)) + ...  ⇒ ∫ sin (x) arcsin (x) dx = ∫(x^2  + (x^6 /(18)) + (x^8 /(30)) + ((2669 x^(10) )/(113400)) + ((601 x^(12) )/(34020))+...)dx                                                      = (x^3 /3) + (x^7 /(126)) + (x^9 /(270)) + ((2669 x^(11) )/(1247400)) + ((601 x^(13) )/(442260)) + ... + k  i still personally think this problem will be cool if we had boundary   conditions  like ∫_0 ^1  sin (x) arcsin (x) dx.

letmegiveatry.sin(x)arcsin(x)dxusingthetaylorseriesexpansionforsin(x)arcsin(x)centredat0sin(x)arcsin(x)=x2+x618+x830+2669x10113400+601x1234020+...sin(x)arcsin(x)dx=(x2+x618+x830+2669x10113400+601x1234020+...)dx=x33+x7126+x9270+2669x111247400+601x13442260+...+kistillpersonallythinkthisproblemwillbecoolifwehadboundaryconditionslike01sin(x)arcsin(x)dx.

Commented by M±th+et£s last updated on 30/Mar/20

thank you sir

thankyousir

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